Chapter 3 · Higher tier

GCSE Maths Fractions (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

Calculate with fractions and mixed numbers fluently, including brackets and negatives
Solve multi-step and reverse fractional-change problems
Decide whether a fraction terminates and convert recurring decimals to fractions
Simplify, multiply, divide, add and subtract algebraic fractions
Solve equations that contain fractions
Prove results involving fractions using algebra

You should already know

Fraction arithmetic from the Foundation chapter
Factorising quadratics and the difference of two squares
Solving linear equations

Where this leads

Rearranging formulae, functions, quadratic equations, algebraic proof, probability and rates of change

Edexcel GCSE (9–1) Mathematics 1MA1, Higher tier (grades 4–9). Content references N2, N8, N10, N12, R3, R9, A4, A5, A17. (Check references against the latest published specification before publishing.)

Printable worksheets. Every lesson in this chapter has a free worksheet and an answer sheet to download and print.

3.1 Fraction arithmetic with confidence: Worksheet (PDF) · Answers (PDF) · lesson page
3.2 Multi-step and reverse fraction problems: Worksheet (PDF) · Answers (PDF) · lesson page
3.3 Recurring decimals and fractions: Worksheet (PDF) · Answers (PDF) · lesson page
3.4 Simplifying algebraic fractions: Worksheet (PDF) · Answers (PDF) · lesson page
3.5 Algebraic fractions and equations: Worksheet (PDF) · Answers (PDF) · lesson page
3.6 Reasoning and proof with fractions: Worksheet (PDF) · Answers (PDF) · lesson page
Check-up and exam-style questions: Worksheet (PDF) · Answers (PDF) · lesson page

Exercise P Do you remember?

Non-calculator

FWork out 34 + 25.

Done

FWork out 23 ÷ 49.

Done

FFactorise x² + 5x + 6.

Done

FFactorise x² − 25.

Done

FSolve 3x + 4 = 19.

Done

3.1

Fraction arithmetic with confidence

Edexcel 1MA1: N2, N8
Grade 4–6

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Fraction arithmetic with confidenceComing soon

At Higher level you are expected to handle fractions quickly and accurately inside longer calculations, with brackets, mixed numbers and negative values. Everything here builds on the four operations from Foundation.

Key ideas

Follow BIDMAS: brackets, indices, then division and multiplication, then addition and subtraction.
Work out the top and bottom of a “big” fraction separately, then divide: ab ÷ cd.
The sign rules for integers apply to fractions: a negative × a negative is positive; a negative ÷ a positive is negative.
Always give the answer in its simplest form, and as a mixed number when the question works with mixed numbers.

Worked example 3.1.1

Work out 315 − 134

Method. To subtract fractions, give them the same denominator, then subtract the numerators and keep the denominator.

Read the question and decide the plan. We can only subtract pieces that are the same size. So first look at the type of numbers and at the denominators (the bottom numbers).
315 − 134
Change the mixed numbers to improper fractions. A mixed number is a whole number plus a fraction. Multiply the whole number by the denominator, add the numerator, and keep the denominator. This lets us calculate with a single fraction.
315 = (3 × 5 + 1) over 5 = 165
134 = (1 × 4 + 3) over 4 = 74
Find a common denominator: the LCM of 5 and 4. We can only add or subtract pieces of the same size, so the denominators must match. The lowest common multiple (LCM) is the smallest number that both denominators divide into.
Multiples of 5: 5, 10, 15, 20
Multiples of 4: 4, 8, 12, 16, 20
LCM = 20
Rewrite both fractions with denominator 20. Multiply the top and the bottom by the same number so the value does not change.
165 = 6420 (× 4)
74 = 3520 (× 5)
Subtract the numerators and keep the denominator. The denominator only tells us the size of each piece, so it stays the same.
6420 − 3520 = 2920
Write the improper fraction as a mixed number. The top is bigger than the bottom, so the answer is more than 1. Divide to find the whole number and the remainder.
29 ÷ 20 = 1 remainder 9
2920 = 1920

Check. A different route, as decimals: 3.2 − 1.75 = 1.45, and our answer 1920 = 1.45. ✓

Watch out. Subtract in the right order: the first fraction minus the second.

Worked example 3.1.2

Work out (23 + 14) ÷ (56 − 12)

Method. Follow BIDMAS: work out each bracket completely first, then divide by flipping the second fraction.

Decide the order of working. BIDMAS says brackets come first. There are two brackets, so each must become a single fraction before we divide.
(23 + 14) ÷ (56 − 12)
Work out the first bracket. To add we need a common denominator. The LCM of 3 and 4 is 12.
23 = 812 and 14 = 312
812 + 312 = 1112
Work out the second bracket. The LCM of 6 and 2 is 6, so write a half as three sixths. Then subtract and simplify.
12 = 36
56 − 36 = 26 = 13
Keep, change, flip. Dividing by a fraction is multiplying by its reciprocal. Flip only the second fraction.
1112 ÷ 13 = 1112 × 31
Cancel and multiply. The 3 on top and the 12 on the bottom share a factor of 3.
1112 × 31 = 114
Write as a mixed number. The top is bigger than the bottom, so divide: 11 ÷ 4 = 2 remainder 3.
114 = 234

Check. A different route, as decimals: 0.667 + 0.25 = 0.917 and 0.833 − 0.5 = 0.333. Then 0.917 ÷ 0.333 = 2.75, and 2 3/4 = 2.75. ✓

Watch out. Do not flip until both brackets are single fractions. Flip only the fraction you are dividing by.

Worked example 3.1.3

Work out 25 − 34

Method. To subtract fractions, give them the same denominator, then subtract the numerators and keep the denominator.

Read the question and decide the plan. We can only subtract pieces that are the same size. So first look at the type of numbers and at the denominators (the bottom numbers).
25 − 34
Find a common denominator: the LCM of 5 and 4. We can only add or subtract pieces of the same size, so the denominators must match. The lowest common multiple (LCM) is the smallest number that both denominators divide into.
Multiples of 5: 5, 10, 15, 20
Multiples of 4: 4, 8, 12, 16, 20
LCM = 20
Rewrite both fractions with denominator 20. Multiply the top and the bottom by the same number so the value does not change.
25 = 820 (× 4)
34 = 1520 (× 5)
Subtract the numerators and keep the denominator. The denominator only tells us the size of each piece, so it stays the same.
820 − 1520 = −720
Write the answer. The result is negative because the second fraction was bigger.
−720

Check. Add back: −720 + 1520 = 820 = 25, the number we started with. ✓

Watch out. The answer is negative because the second fraction is bigger than the first. Keep the order: first minus second.

Worked example 3.1.4

Work out 1 ÷ (12 + 13)

Method. Work out the bracket first. Then dividing 1 by a fraction gives its reciprocal.

Do the bracket first. BIDMAS says brackets come before division.
12 + 13
Find a common denominator. The LCM of 2 and 3 is 6.
12 = 36 (× 3) and 13 = 26 (× 2)
Add the numerators. The pieces are now the same size, so keep the denominator.
36 + 26 = 56
Divide 1 by the result. Write 1 as a fraction over 1, then keep, change, flip.
1 ÷ 56 = 11 × 65 = 65
Write as a mixed number. 6 ÷ 5 = 1 remainder 1.
65 = 115

Check. As decimals: 0.5 + 0.333 = 0.833 and 1 ÷ 0.833 = 1.2, which is 115. ✓

Watch out. 11 ÷ (12 + 13) is not 11 ÷ 12 + 11 ÷ 13. You cannot split a division over a sum.

Worked example 3.1.5

Work out −58 ÷ −1516

Method. Decide the sign first with the rules for negatives, then divide the fractions as normal.

Decide the sign of the answer. A negative divided by a negative is positive. So we can work with the positive values and the answer is positive.
− ÷ − = +, so we work out 58 ÷ 1516
Keep, change, flip. Turn the second fraction upside down and change the division to a multiplication.
58 ÷ 1516 = 58 × 1615
Cancel common factors. 5 on the top and 15 on the bottom share a factor of 5. 16 on the top and 8 on the bottom share a factor of 8.
58 × 1615 = 11 × 23
Multiply across. Top times top and bottom times bottom.
Tops: 1 × 2 = 2
Bottoms: 1 × 3 = 3
23
Write the answer with its sign. The sign was decided in step 1: positive.
23

Check. Multiply back: 23 × −1516 = −3048 = −58, which is the number we started with. ✓

Watch out. Two negatives in a division give a positive. Do not put a minus sign in the answer.

Watch out

In (23 + 14) ÷ (56 − 12), do not flip until both brackets are single fractions. Flip only the fraction you are dividing by.

3.2

Multi-step and reverse fraction problems

Edexcel 1MA1: N12, R3, R9
Grade 5–7

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Multi-step and reverse fraction problemsComing soon

Many exam questions describe a story in several steps. The key is to track the fraction of the original that is left at each stage, and to work backwards when the final amount is known.

Key ideas

Fractional change: increasing by 15 means multiplying by 65. Decreasing by 14 means multiplying by 34.
Successive changes: multiply the multipliers. Do not add the fractions.
Reverse problems: write final = multiplier × original, then divide the final amount by the multiplier.
Part and remainder: “13 of the rest” means 13 of the remaining amount, not of the original.

Worked example 3.2.1

A price is reduced by 15 and then reduced by 14 of the new price. The final price is £90. Find the original price.

Method. Turn each change into a multiplier, multiply the multipliers, then work backwards from the final price.

Find the first multiplier. Reducing by 15 leaves 45 of the price.
1 − 15 = 45
Find the second multiplier. The second reduction is 14 of the new price, so 34 of the new price is left.
1 − 14 = 34
Find the overall multiplier. The second change applies to the result of the first, so multiply the multipliers (do not add the fractions). Cancel the 4s.
45 × 34 = 35
Link the final price to the original. The final price is 35 of the original. 3 parts are worth £90, so one part (one fifth) is worth £90 ÷ 3.
35 of original = £90
1 fifth = £90 ÷ 3 = £30
Find the original price. The whole is 5 fifths.
5 × £30 = £150

Check. Work forwards from £150. The first reduction leaves 4/5 of £150 = £120. The second leaves 3/4 of £120 = £90. This matches. ✓

Watch out. Do not add the fractions. A reduction of 15 then 14 is not a reduction of 920, because the second reduction is of a smaller price.

Worked example 3.2.2

In a box of chocolates, 38 are milk. 13 of the rest are plain. The other 30 chocolates are white. How many chocolates are in the box?

Method. Find the fraction of the whole box that is white, then use the 30 white chocolates to find the whole.

Find the fraction that is not milk. The rest means everything except the milk chocolates.
1 − 38 = 58
Find the fraction that is plain. Plain is 13 of the rest, which is 13 of 58. “Of” means multiply.
13 × 58 = 524
Find the fraction that is white. White is what is left after the milk and the plain. Use a common denominator of 24.
58 − 524 = 1524 − 524 = 1024 = 512
Link the white chocolates to the whole box. 512 of the box is 30 chocolates. 5 parts are 30, so one twelfth is 30 ÷ 5.
512 of the box = 30
1 twelfth = 30 ÷ 5 = 6
Find the whole box. The whole is 12 twelfths.
12 × 6 = 72 chocolates

Check. Work forwards from 72. Milk: 38 of 72 = 27. The rest is 45. Plain: 13 of 45 = 15. White: 45 − 15 = 30. This matches. ✓

Watch out. “Of the rest” means of the amount that remains after the first group, not of the whole box.

Worked example 3.2.3

Pipe A fills a tank in 6 hours. Pipe B fills it in 3 hours. How long do they take together?

Method. For work-rate problems, find the fraction of the job each does in one hour, add them, then turn the combined rate into a time.

Find how much A fills in one hour. A fills the whole tank in 6 hours, so in 1 hour it fills one sixth.
A: 16 of the tank per hour
Find how much B fills in one hour. B fills the whole tank in 3 hours, so in 1 hour it fills one third.
B: 13 of the tank per hour
Add the rates. Working together they fill both amounts every hour. Use a common denominator of 6.
16 + 26 = 36 = 12 of the tank per hour
Turn the combined rate into a time. If half the tank is filled each hour, the whole tank takes 1 ÷ 12 = 2 hours.
1 ÷ 12 = 2
Write the answer with units. The time is in hours.
2 hours

Check. In 2 hours A fills 2 × 16 = 13 and B fills 2 × 13 = 23. 13 + 23 = 1 whole tank. ✓

Watch out. Do not just average the two times, and do not add 6 + 3. Add the rates (fractions per hour), then take the reciprocal.

Worked example 3.2.4

A price of £100 is increased by 15 and then the new price is decreased by 15. What is the final price?

Method. Turn each change into a multiplier and apply them one after the other. The second change is a fraction of the new price.

Find the multiplier for the increase. Increasing by 15 means adding a fifth, so the price becomes 1 + 15 of what it was.
1 + 15 = 65
Apply the increase. Divide by the denominator 5, then multiply by 6.
£100 ÷ 5 = £20
£20 × 6 = £120
Find the multiplier for the decrease. Decreasing by 15 of the new price leaves 1 − 15 of it.
1 − 15 = 45
Apply the decrease. Work on the new price, £120, not on the original £100.
£120 ÷ 5 = £24
£24 × 4 = £96

Check. A different route: the overall multiplier is 65 × 45 = 2425. 2425 of £100 = £96. ✓

Watch out. The price does not return to £100. The decrease is of a bigger price, so it takes off more than the increase added.

Worked example 3.2.5

A bike is reduced by 27 in a sale. The sale price is £150. What was the original price?

Method. Reverse problem: write sale price = original × multiplier, then divide the sale price by the multiplier.

Find the multiplier. A reduction of 27 leaves 57 of the original price.
1 − 27 = 57
Write the relationship. The sale price is the multiplier times the original price.
57 of original = £150
Find one part. 5 parts are £150, so one seventh is £150 ÷ 5.
£150 ÷ 5 = £30 for one seventh
Find the whole. The original price is 7 sevenths.
£30 × 7 = £210

Check. Work forwards: 27 of £210 = £60 reduction, and £210 − £60 = £150. ✓

Watch out. Do not increase £150 by 27. The reduction was a fraction of the original, so you must divide by the multiplier.

Watch out

Increasing a price by 14 and then decreasing by 14 does not return to the start: 54 × 34 = 1516, a loss of 116.

3.3

Recurring decimals and fractions

Edexcel 1MA1: N10
Grade 5–8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Recurring decimals and fractionsComing soon

Every fraction is either a terminating decimal (it ends, like 38 = 0.375) or a recurring decimal (a pattern repeats forever, like 13 = 0.333…). We show recurring digits with a dot above the first and last digit of the repeating block: 0.3, 0.18, 0.53.

Key ideas

Terminating test: write the fraction in simplest form. It terminates if (and only if) the denominator has no prime factors other than 2 and 5.
Fraction → decimal: divide the numerator by the denominator.
Recurring decimal → fraction: let x be the decimal, multiply by a power of 10 to line the repeating parts up, subtract to cancel them, then solve for x.

Worked example 3.3.1

Does 730 or 740 give a terminating decimal?

Method. A fraction in simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5.

Say what the test is. A terminating decimal is one that ends. Only denominators made from 2s and 5s can be changed to tenths, hundredths and so on, because 10 = 2 × 5.
Look for prime factors other than 2 and 5
Check both fractions are in simplest form. 7 is prime, and it does not divide into 30 or 40, so neither fraction can be simplified.
730 and 740 are already simplest
Break the denominator of 730 into prime factors. Divide by primes until we reach 1.
30 = 2 × 3 × 5
There is a factor of 3, which is not 2 or 5.
Break the denominator of 740 into prime factors. Do the same for 40.
40 = 2 × 2 × 2 × 5 = 2³ × 5
Only 2s and 5s.
Write the conclusion. Answer the question that was asked.
740 terminates. 730 does not terminate (it recurs).

Check. A different route: divide. 740 = 0.175, which ends. 730 = 0.2333…, where the 3 repeats for ever. ✓

Watch out. A factor of 3 in the denominator of a fraction in simplest form always gives a recurring decimal. But always simplify first: 615 = 25 terminates.

Worked example 3.3.2

Write 512 as a decimal, using dot notation for the repeating digits.

Method. A fraction whose denominator has a prime factor other than 2 or 5 gives a recurring decimal. Use long division and watch for a remainder that repeats.

Say what the fraction means. The fraction bar means divide. The denominator has a prime factor other than 2 or 5, so we expect the decimal to repeat.
512 = 5 ÷ 12
Set up the long division 5 ÷ 12. Write the number with a decimal point and extra zeros so we can keep dividing. Adding zeros after the point does not change its value.
5.000000… ÷ 12
Divide step by step. At each step carry the remainder across and put a zero after it. Keep going until a remainder appears for a second time.
12 into 5 goes 0 times, remainder 5
12 into 50 goes 4 times, remainder 2
12 into 20 goes 1 time, remainder 8
12 into 80 goes 6 times, remainder 8
The remainder 8 has appeared before, so the digits from that point repeat for ever.
Find the repeating block. When a remainder comes back, the same working happens again, so the same digits repeat. The repeating block is the digits that were found between the two matching remainders.
Digits found: 0.416
Repeating block: 6
Non-repeating digits: 41
Write the answer with dot notation. Put a dot over the first digit and the last digit of the repeating block (just one dot if only one digit repeats).
512 = 0.416

Check. A different check: multiply the decimal by the denominator. Using the first few digits, 0.41666667 × 12 = 5.00000000, which is extremely close to 5. ✓

Watch out. Show the dots over the first and the last digit of the block. Writing only a few digits and “…” is not enough to show which digits repeat.

Worked example 3.3.3

Write 0.7 as a fraction in its simplest form.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
0.7 = 0.777…
Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
x = 0.7 = 0.777…
Multiply by 10. 1 digit repeats, so multiplying by 10¹ = 10 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
10x = 7.77…
Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
10x − x = 7.77… − 0.77…
9x = 7
Solve for x. Divide both sides by the number in front of x.
x = 79
Check the fraction is in its simplest form. 7 and 9 have no common factor except 1.
x = 79

Check. A different route: divide the top by the bottom. 7 ÷ 9 gives 0.7, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Worked example 3.3.4

Write 0.45 as a fraction in its simplest form.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
0.45 = 0.454545…
Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
x = 0.45 = 0.454545…
Multiply by 100. 2 digits repeat, so multiplying by 10² = 100 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
100x = 45.4545…
Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
100x − x = 45.4545… − 0.4545…
99x = 45
Solve for x. Divide both sides by the number in front of x.
x = 4599
Simplify the fraction. The highest common factor of 45 and 99 is 9, so divide the top and the bottom by 9.
4599 = 511

Check. A different route: divide the top by the bottom. 5 ÷ 11 gives 0.45, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Worked example 3.3.5

Write 0.27 as a fraction in its simplest form.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
0.27 = 0.2777…
Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
x = 0.27 = 0.2777…
Multiply by 10. This moves the non-repeating digit (2) to the left of the point so that the repeating digits start straight after the point.
10x = 2.77…
Multiply by 100. This moves one more full block of repeating digits to the left of the point, so both lines have the same tail.
100x = 27.77…
Subtract. Both lines have the same repeating tail after the point, so the tail cancels.
100x − 10x = 27 − 2
90x = 25
Solve for x. Divide both sides by the number in front of x.
x = 2590
Simplify the fraction. The highest common factor of 25 and 90 is 5, so divide the top and the bottom by 5.
2590 = 518

Check. A different route: divide the top by the bottom. 5 ÷ 18 gives 0.27, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Exam tip

To check a recurring decimal answer, divide the numerator by the denominator on a calculator, or use short division by hand.

3.4

Simplifying algebraic fractions

Edexcel 1MA1: A4, A5
Grade 6–8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Simplifying algebraic fractionsComing soon

An algebraic fraction has algebra in the numerator, denominator or both. The rules are exactly the same as for number fractions: simplify by dividing top and bottom by a common factor.

Key ideas

Factorise first. Then cancel factors that appear in both the numerator and the denominator.
You can cancel factors (things multiplied), never terms (things added). x + 33 cannot be simplified by cancelling the 3s.
Multiply: multiply numerators, multiply denominators, cancelling common factors first.
Divide: keep the first fraction, flip the second, then multiply.

Worked example 3.4.1

Simplify 6x²y9xy

Method. To simplify an algebraic fraction, write the top and the bottom as products, then cancel the numbers and each letter that appears on both.

Write the top and the bottom as multiplications. A power like x² means x × x. Writing everything as a product lets us see every factor.
6x²y = 6 × x × x × y
9xy = 9 × x × y
Cancel the numbers. The number parts are 6 and 9. Their highest common factor is 3: 6 ÷ 3 = 2 and 9 ÷ 3 = 3.
69 = 23
Cancel one x. One x on the bottom cancels with one of the two xs on the top, leaving one x on the top.
x × x on top, x on the bottom
Cancel the y. The one y on the top cancels with the one y on the bottom.
y on top, y on the bottom
Write what is left. Multiply together everything that has not been crossed out.
2x3

Check. Try x = 3 and y = 2. Original: top = 6 × 9 × 2 = 108, bottom = 9 × 3 × 2 = 54, so 108 ÷ 54 = 2. Answer: 2 × 3 ÷ 3 = 2. They match. ✓

Watch out. Cancel numbers with numbers and each letter with the same letter. You can only cancel things that are multiplied, never things that are added.

Worked example 3.4.2

Simplify x² + 5xx² − 25

Key words. A factor is something that is multiplied. For example 2 and 3 are factors of 6, and (x + 1) is a factor of 3(x + 1). Factorise means rewrite as a product. Cancel means divide the top and the bottom of a fraction by the same factor. An algebraic fraction is just a fraction with letters in it, and it follows exactly the same rules as number fractions.

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
x² + 5xx² − 25
Factorise the top: x² + 5x Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Look for a common factor. A common factor is something that divides into every term.
The terms are x², 5x. The biggest number and the highest power of x that divide into all of them is x.
Take it outside a bracket, and inside the bracket put each term divided by it:
x² ÷ x = x
5x ÷ x = 5
So x² + 5x = x(x + 5).
Check by expanding: x × (x + 5) = x × x, x × 5 → x² + 5x. ✓
x² + 5x = x(x + 5)
Factorise the bottom: x² − 25 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 25 = 5², so b = 5.
Put them into the rule: (x − 5)(x + 5).
Check by expanding: (x − 5)(x + 5) = x² + 5x − 5x − 25 = x² − 25. ✓ (the middle terms cancel)
x² − 25 = (x − 5)(x + 5)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
x(x + 5)(x − 5)(x + 5)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x + 5).
x~~(x + 5)~~(x − 5)~~(x + 5)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
x(x − 5)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 14, bottom = -21, so the fraction = -2/3. Our answer at x = 2 = -2/3. They match. ✓

Watch out. You can only cancel factors (things multiplied), never terms (things added). In (x + 6)/x you cannot cancel the x, because the x is added to 6 on top. That is why we always factorise first.

Worked example 3.4.3

Simplify x² + 7x + 10x² − 4

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
x² + 7x + 10x² − 4
Factorise the top: x² + 7x + 10 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 7 and c = 10.
We need two numbers that multiply to 10 (the last number) and add to 7 (the middle number). List the pairs that multiply to 10:
1 × 10 = 10, add: 1 + 10 = 11 ✗
-1 × -10 = 10, add: -1 + -10 = -11 ✗
2 × 5 = 10, add: 2 + 5 = 7 ✓ this pair works
Put the pair into two brackets: (x + 2)(x + 5).
Check by expanding: x² + 5x + 2x + 10 = x² + 7x + 10. ✓
x² + 7x + 10 = (x + 2)(x + 5)
Factorise the bottom: x² − 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 4 = 2², so b = 2.
Put them into the rule: (x − 2)(x + 2).
Check by expanding: (x − 2)(x + 2) = x² + 2x − 2x − 4 = x² − 4. ✓ (the middle terms cancel)
x² − 4 = (x − 2)(x + 2)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
(x + 2)(x + 5)(x − 2)(x + 2)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x + 2).
~~(x + 2)~~(x + 5)(x − 2)~~(x + 2)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
(x + 5)(x − 2)

Check. Choose a number for x that does not make any bottom zero. Try x = 3. The original: top = 40, bottom = 5, so the fraction = 8. Our answer at x = 3 = 8. They match. ✓

Worked example 3.4.4

Work out x² − 1x + 2 ÷ x + 1x + 2

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal, which is the fraction turned upside down. Here is why it works with numbers: 1/2 ÷ 1/4 asks "how many quarters fit into a half?" The answer is 2, and 1/2 × 4/1 = 2. So: keep the first fraction, change ÷ to ×, and flip the second fraction.
x² − 1x + 2 ÷ x + 1x + 2 = x² − 1x + 2 × x + 2x + 1
Factorise the first top: x² − 1 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 1 = 1², so b = 1.
Put them into the rule: (x − 1)(x + 1).
Check by expanding: (x − 1)(x + 1) = x² + 1x − 1x − 1 = x² − 1. ✓ (the middle terms cancel)
x² − 1 = (x − 1)(x + 1)
Factorise the first bottom: x + 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 2 is a single bracket already. It cannot be factorised any further.
x + 2 = (x + 2)
Factorise the second bottom: x + 1 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 1 is a single bracket already. It cannot be factorised any further.
x + 1 = (x + 1)
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
(x − 1)(x + 1)(x + 2) × (x + 2)(x + 1)
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
(x − 1)(x + 1) × (x + 2)(x + 2) × (x + 1)
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x + 1), (x + 2).
(x − 1)~~(x + 1)(x + 2)(x + 1)(x + 2)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
(x − 1)

Check. Try x = 2 (it does not make any bottom zero). The original expression gives 1. Our answer gives 1. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

Worked example 3.4.5

Work out x² − 9x + 4 × x + 4x + 3

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Factorise the first top: x² − 9 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 9 = 3², so b = 3.
Put them into the rule: (x − 3)(x + 3).
Check by expanding: (x − 3)(x + 3) = x² + 3x − 3x − 9 = x² − 9. ✓ (the middle terms cancel)
x² − 9 = (x − 3)(x + 3)
Factorise the first bottom: x + 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 4 is a single bracket already. It cannot be factorised any further.
x + 4 = (x + 4)
Factorise the second bottom: x + 3 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 3 is a single bracket already. It cannot be factorised any further.
x + 3 = (x + 3)
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
(x − 3)(x + 3)(x + 4) × (x + 4)(x + 3)
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
(x − 3)(x + 3) × (x + 4)(x + 4) × (x + 3)
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x + 3), (x + 4).
(x − 3)~~(x + 3)(x + 4)(x + 3)(x + 4)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
(x − 3)

Check. Try x = 2 (it does not make any bottom zero). The original expression gives -1. Our answer gives -1. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

Watch out

Do not cancel across addition. x + 6x is not equal to 7, and 2x + 42 = x + 2, not 2x + 2. Cancel only after factorising: 2(x + 2) ÷ 2.

Exercise 3D Simplifying, multiplying and dividing algebraic fractions

Non-calculator

FSimplify.

(a)6x9

(b)12x²8x

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
12x²8x
Write the top as a product: 12x² A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
12x² = 12 × x × x
Write the bottom as a product: 8x A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
8x = 8 × x
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
12x²8x
Cancel the numbers first. The number part of the top is 12 and the number part of the bottom is 8. They have a common factor of 4, so divide both number parts by 4. (This is the same as simplifying the fraction 128 = 32.)
12x²8x = 3x²2x
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: x.
3xx2x
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
3x2

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 48, bottom = 16, so the fraction = 3. Our answer at x = 2 = 3. They match. ✓

(d)3x + 69

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
3x + 69
Factorise the top: 3x + 6 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Look for a common factor. A common factor is something that divides into every term.
The terms are 3x, 6. The biggest number and the highest power of x that divide into all of them is 3.
Take it outside a bracket, and inside the bracket put each term divided by it:
3x ÷ 3 = x
6 ÷ 3 = 2
So 3x + 6 = 3(x + 2).
Check by expanding: 3 × (x + 2) = 3 × x, 3 × 2 → 3x + 6. ✓
3x + 6 = 3(x + 2)
Write the bottom as a product: 9 A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
9 = 9
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
3(x + 2)9
Cancel the numbers first. The number part of the top is 3 and the number part of the bottom is 9. They have a common factor of 3, so divide both number parts by 3. (This is the same as simplifying the fraction 39 = 13.)
3(x + 2)9 = (x + 2)3
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
1/3(x + 2)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 12, bottom = 9, so the fraction = 4/3. Our answer at x = 2 = 4/3. They match. ✓

Done

FFactorise and simplify fully.

(a)x² + 4xx + 4

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
x² + 4xx + 4
Factorise the top: x² + 4x Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Look for a common factor. A common factor is something that divides into every term.
The terms are x², 4x. The biggest number and the highest power of x that divide into all of them is x.
Take it outside a bracket, and inside the bracket put each term divided by it:
x² ÷ x = x
4x ÷ x = 4
So x² + 4x = x(x + 4).
Check by expanding: x × (x + 4) = x × x, x × 4 → x² + 4x. ✓
x² + 4x = x(x + 4)
Factorise the bottom: x + 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 4 is a single bracket already. It cannot be factorised any further.
x + 4 = (x + 4)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
x(x + 4)(x + 4)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x + 4).
x~~(x + 4)(x + 4)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
x

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 12, bottom = 6, so the fraction = 2. Our answer at x = 2 = 2. They match. ✓

(b)x² − 9x + 3

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
x² − 9x + 3
Factorise the top: x² − 9 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 9 = 3², so b = 3.
Put them into the rule: (x − 3)(x + 3).
Check by expanding: (x − 3)(x + 3) = x² + 3x − 3x − 9 = x² − 9. ✓ (the middle terms cancel)
x² − 9 = (x − 3)(x + 3)
Factorise the bottom: x + 3 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 3 is a single bracket already. It cannot be factorised any further.
x + 3 = (x + 3)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
(x − 3)(x + 3)(x + 3)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x + 3).
(x − 3)~~(x + 3)(x + 3)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
(x − 3)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = -5, bottom = 5, so the fraction = -1. Our answer at x = 2 = -1. They match. ✓

(c)x² + 5x + 6x + 2

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
x² + 5x + 6x + 2
Factorise the top: x² + 5x + 6 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 5 and c = 6.
We need two numbers that multiply to 6 (the last number) and add to 5 (the middle number). List the pairs that multiply to 6:
1 × 6 = 6, add: 1 + 6 = 7 ✗
-1 × -6 = 6, add: -1 + -6 = -7 ✗
2 × 3 = 6, add: 2 + 3 = 5 ✓ this pair works
Put the pair into two brackets: (x + 2)(x + 3).
Check by expanding: x² + 3x + 2x + 6 = x² + 5x + 6. ✓
x² + 5x + 6 = (x + 2)(x + 3)
Factorise the bottom: x + 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 2 is a single bracket already. It cannot be factorised any further.
x + 2 = (x + 2)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
(x + 2)(x + 3)(x + 2)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x + 2).
~~(x + 2)~~(x + 3)~~(x + 2)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
(x + 3)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 20, bottom = 4, so the fraction = 5. Our answer at x = 2 = 5. They match. ✓

(d)2x² − 8x² − 4x + 4

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
2x² − 8x² − 4x + 4
Factorise the top: 2x² − 8 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Look for a common factor. A common factor is something that divides into every term.
The terms are 2x², −8. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x² ÷ 2 = x²
−8 ÷ 2 = −4
So 2x² − 8 = 2(x² − 4).
Check by expanding: 2 × (x² − 4) = 2 × x², 2 × (−4) → 2x² − 8. ✓
Can the bracket be factorised again? Look at (x² − 4).
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 4 = 2², so b = 2.
Put them into the rule: (x − 2)(x + 2).
Check by expanding: (x − 2)(x + 2) = x² + 2x − 2x − 4 = x² − 4. ✓ (the middle terms cancel)
2x² − 8 = 2(x − 2)(x + 2)
Factorise the bottom: x² − 4x + 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -4 and c = 4.
We need two numbers that multiply to 4 (the last number) and add to -4 (the middle number). List the pairs that multiply to 4:
1 × 4 = 4, add: 1 + 4 = 5 ✗
-1 × -4 = 4, add: -1 + -4 = -5 ✗
2 × 2 = 4, add: 2 + 2 = 4 ✗
-2 × -2 = 4, add: -2 + -2 = -4 ✓ this pair works
Put the pair into two brackets: (x − 2)(x − 2).
Check by expanding: x² − 2x − 2x + 4 = x² − 4x + 4. ✓
x² − 4x + 4 = (x − 2)²
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
2(x − 2)(x + 2)(x − 2)²
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x − 2).
2~~(x − 2)~~(x + 2)~~(x − 2)~~(x − 2)
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
2(x + 2)(x − 2)

Check. Choose a number for x that does not make any bottom zero. Try x = 3. The original: top = 10, bottom = 1, so the fraction = 10. Our answer at x = 3 = 10. They match. ✓

(e)x² − x − 12x² − 16

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
x² − x − 12x² − 16
Factorise the top: x² − x − 12 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -1 and c = -12.
We need two numbers that multiply to -12 (the last number) and add to -1 (the middle number). List the pairs that multiply to -12:
1 × -12 = -12, add: 1 + -12 = -11 ✗
-1 × 12 = -12, add: -1 + 12 = 11 ✗
2 × -6 = -12, add: 2 + -6 = -4 ✗
-2 × 6 = -12, add: -2 + 6 = 4 ✗
3 × -4 = -12, add: 3 + -4 = -1 ✓ this pair works
Put the pair into two brackets: (x + 3)(x − 4).
Check by expanding: x² − 4x + 3x − 12 = x² − x − 12. ✓
x² − x − 12 = (x − 4)(x + 3)
Factorise the bottom: x² − 16 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 16 = 4², so b = 4.
Put them into the rule: (x − 4)(x + 4).
Check by expanding: (x − 4)(x + 4) = x² + 4x − 4x − 16 = x² − 16. ✓ (the middle terms cancel)
x² − 16 = (x − 4)(x + 4)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
(x − 4)(x + 3)(x − 4)(x + 4)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x − 4).
~~(x − 4)~~(x + 3)~~(x − 4)~~(x + 4)
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
(x + 3)(x + 4)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = -10, bottom = -12, so the fraction = 5/6. Our answer at x = 2 = 5/6. They match. ✓

(f)3x² + x − 2x² − 1

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
3x² + x − 2x² − 1
Factorise the top: 3x² + x − 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. A quadratic where the number in front of x² is 3, not 1. We use the split the middle term method.
Multiply the first and last numbers: 3 × (-2) = -6. We need two numbers that multiply to -6 and add to 1 (the middle number).
1 × -6 = -6, add: -5 ✗
-1 × 6 = -6, add: 5 ✗
2 × -3 = -6, add: -1 ✗
-2 × 3 = -6, add: 1 ✓ this pair works
Split the middle term (1x) into -2x and 3x: 3x² − 2x + 3x − 2.
Factorise each pair separately. First pair: 3x² − 2x = x(3x − 2). Second pair: 3x − 2 = 1(3x − 2).
The two brackets (3x − 2) are the same. That is how we know it has worked. Write: x(3x − 2) + 1(3x − 2).
Take the common bracket out: (3x − 2) multiplied by (x + 1).
Result: (x + 1)(3x − 2).
Check by expanding: (x + 1)(3x − 2) = 3x² + x − 2. ✓
3x² + x − 2 = (x + 1)(3x − 2)
Factorise the bottom: x² − 1 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 1 = 1², so b = 1.
Put them into the rule: (x − 1)(x + 1).
Check by expanding: (x − 1)(x + 1) = x² + 1x − 1x − 1 = x² − 1. ✓ (the middle terms cancel)
x² − 1 = (x − 1)(x + 1)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
(x + 1)(3x − 2)(x − 1)(x + 1)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x + 1).
~~(x + 1)~~(3x − 2)(x − 1)~~(x + 1)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
(3x − 2)(x − 1)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 12, bottom = 3, so the fraction = 4. Our answer at x = 2 = 4. They match. ✓

Done

FWork out. Give each answer in its simplest form.

(a)x4 × 8x²

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Write the first top as a product: x A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
x = x
Write the first bottom as a product: 4 A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
4 = 4
Write the second top as a product: 8 A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
8 = 8
Write the second bottom as a product: x² A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
x² = x × x
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
x4 × 8x²
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
x × 84 × x²
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: x.
8x4xx
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
2x

Check. Try x = 2 (it does not make any bottom zero). The original expression gives 1. Our answer gives 1. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

(b)3x5 ÷ 9x10

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal, which is the fraction turned upside down. Here is why it works with numbers: 1/2 ÷ 1/4 asks "how many quarters fit into a half?" The answer is 2, and 1/2 × 4/1 = 2. So: keep the first fraction, change ÷ to ×, and flip the second fraction.
3x5 ÷ 9x10 = 3x5 × 109x
Write the first top as a product: 3x A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
3x = 3 × x
Write the first bottom as a product: 5 A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
5 = 5
Write the second top as a product: 10 A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
10 = 10
Write the second bottom as a product: 9x A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
9x = 9 × x
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
3x5 × 109x
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
3x × 105 × 9x
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: x.
30x45x
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
23

Check. Try x = 2 (it does not make any bottom zero). The original expression gives 2/3. Our answer gives 2/3. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

(c)x + 3x − 2 × x − 2x + 5

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Factorise the first top: x + 3 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 3 is a single bracket already. It cannot be factorised any further.
x + 3 = (x + 3)
Factorise the first bottom: x − 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x − 2 is a single bracket already. It cannot be factorised any further.
x − 2 = (x − 2)
Factorise the second bottom: x + 5 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 5 is a single bracket already. It cannot be factorised any further.
x + 5 = (x + 5)
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
(x + 3)(x − 2) × (x − 2)(x + 5)
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
(x + 3) × (x − 2)(x − 2) × (x + 5)
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x − 2).
~~(x − 2)~~(x + 3)~~(x − 2)~~(x + 5)
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
(x + 3)(x + 5)

Check. Try x = 3 (it does not make any bottom zero). The original expression gives 3/4. Our answer gives 3/4. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

(d)x² − 1x + 2 ÷ x + 1x + 2

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal, which is the fraction turned upside down. Here is why it works with numbers: 1/2 ÷ 1/4 asks "how many quarters fit into a half?" The answer is 2, and 1/2 × 4/1 = 2. So: keep the first fraction, change ÷ to ×, and flip the second fraction.
x² − 1x + 2 ÷ x + 1x + 2 = x² − 1x + 2 × x + 2x + 1
Factorise the first top: x² − 1 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 1 = 1², so b = 1.
Put them into the rule: (x − 1)(x + 1).
Check by expanding: (x − 1)(x + 1) = x² + 1x − 1x − 1 = x² − 1. ✓ (the middle terms cancel)
x² − 1 = (x − 1)(x + 1)
Factorise the first bottom: x + 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 2 is a single bracket already. It cannot be factorised any further.
x + 2 = (x + 2)
Factorise the second bottom: x + 1 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 1 is a single bracket already. It cannot be factorised any further.
x + 1 = (x + 1)
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
(x − 1)(x + 1)(x + 2) × (x + 2)(x + 1)
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
(x − 1)(x + 1) × (x + 2)(x + 2) × (x + 1)
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x + 1), (x + 2).
(x − 1)~~(x + 1)(x + 2)(x + 1)(x + 2)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
(x − 1)

Check. Try x = 2 (it does not make any bottom zero). The original expression gives 1. Our answer gives 1. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

(e)2x + 4x² − 9 × x − 3x + 2

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Factorise the first top: 2x + 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Look for a common factor. A common factor is something that divides into every term.
The terms are 2x, 4. The biggest number and the highest power of x that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2x ÷ 2 = x
4 ÷ 2 = 2
So 2x + 4 = 2(x + 2).
Check by expanding: 2 × (x + 2) = 2 × x, 2 × 2 → 2x + 4. ✓
2x + 4 = 2(x + 2)
Factorise the first bottom: x² − 9 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 9 = 3², so b = 3.
Put them into the rule: (x − 3)(x + 3).
Check by expanding: (x − 3)(x + 3) = x² + 3x − 3x − 9 = x² − 9. ✓ (the middle terms cancel)
x² − 9 = (x − 3)(x + 3)
Factorise the second top: x − 3 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x − 3 is a single bracket already. It cannot be factorised any further.
x − 3 = (x − 3)
Factorise the second bottom: x + 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 2 is a single bracket already. It cannot be factorised any further.
x + 2 = (x + 2)
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
2(x + 2)(x − 3)(x + 3) × (x − 3)(x + 2)
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
2(x + 2) × (x − 3)(x − 3)(x + 3) × (x + 2)
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x − 3), (x + 2).
2~~(x − 3)(x + 2)(x − 3)(x + 2)~~(x + 3)
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
2(x + 3)

Check. Try x = 2 (it does not make any bottom zero). The original expression gives 2/5. Our answer gives 2/5. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

(f)x² + 3xx² − 1 ÷ xx − 1

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal, which is the fraction turned upside down. Here is why it works with numbers: 1/2 ÷ 1/4 asks "how many quarters fit into a half?" The answer is 2, and 1/2 × 4/1 = 2. So: keep the first fraction, change ÷ to ×, and flip the second fraction.
x² + 3xx² − 1 ÷ xx − 1 = x² + 3xx² − 1 × x − 1x
Factorise the first top: x² + 3x Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Look for a common factor. A common factor is something that divides into every term.
The terms are x², 3x. The biggest number and the highest power of x that divide into all of them is x.
Take it outside a bracket, and inside the bracket put each term divided by it:
x² ÷ x = x
3x ÷ x = 3
So x² + 3x = x(x + 3).
Check by expanding: x × (x + 3) = x × x, x × 3 → x² + 3x. ✓
x² + 3x = x(x + 3)
Factorise the first bottom: x² − 1 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 1 = 1², so b = 1.
Put them into the rule: (x − 1)(x + 1).
Check by expanding: (x − 1)(x + 1) = x² + 1x − 1x − 1 = x² − 1. ✓ (the middle terms cancel)
x² − 1 = (x − 1)(x + 1)
Factorise the second top: x − 1 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x − 1 is a single bracket already. It cannot be factorised any further.
x − 1 = (x − 1)
Write the second bottom as a product: x A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
x = x
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
x(x + 3)(x − 1)(x + 1) × (x − 1)x
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
x(x + 3) × (x − 1)(x − 1)(x + 1) × x
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x − 1), x.
~~(x − 1)~~x(x + 3)~~(x − 1)~~x(x + 1)
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
(x + 3)(x + 1)

Check. Try x = 2 (it does not make any bottom zero). The original expression gives 5/3. Our answer gives 5/3. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

Done

RRia says x + 6x simplifies to 7. Explain her mistake.

Done

PSThe area of a rectangle is (x² + 5x + 6) cm² and its length is (x + 3) cm. Find an expression for its width.

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

Decide the operation. Area of a rectangle = length × width, so width = area ÷ length.
Width = x² + 5x + 6x + 3
Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
x² + 5x + 6x + 3
Factorise the top: x² + 5x + 6 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 5 and c = 6.
We need two numbers that multiply to 6 (the last number) and add to 5 (the middle number). List the pairs that multiply to 6:
1 × 6 = 6, add: 1 + 6 = 7 ✗
-1 × -6 = 6, add: -1 + -6 = -7 ✗
2 × 3 = 6, add: 2 + 3 = 5 ✓ this pair works
Put the pair into two brackets: (x + 2)(x + 3).
Check by expanding: x² + 3x + 2x + 6 = x² + 5x + 6. ✓
x² + 5x + 6 = (x + 2)(x + 3)
Factorise the bottom: x + 3 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 3 is a single bracket already. It cannot be factorised any further.
x + 3 = (x + 3)
Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
(x + 2)(x + 3)(x + 3)
Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
Matching factors: (x + 3).
(x + 2)~~(x + 3)(x + 3)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together whatever has not been crossed out.
(x + 2)
Write the answer with units. The width is a length, so it is in centimetres.
Width = (x + 2) cm

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 20, bottom = 5, so the fraction = 4. Our answer at x = 2 = 4. They match. ✓

Done

RShow that x² − 4x + 2 × xx − 2 simplifies to x.

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Plan the proof. In a “show that” question the answer is given, so every step must be shown. Start from one side and use correct algebra until you reach the other side.
Start from the left side: x² − 4x + 2 × xx − 2
Factorise the first top: x² − 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 4 = 2², so b = 2.
Put them into the rule: (x − 2)(x + 2).
Check by expanding: (x − 2)(x + 2) = x² + 2x − 2x − 4 = x² − 4. ✓ (the middle terms cancel)
x² − 4 = (x − 2)(x + 2)
Factorise the first bottom: x + 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 2 is a single bracket already. It cannot be factorised any further.
x + 2 = (x + 2)
Write the second top as a product: x A single term is already a product (everything is multiplied together). Writing out the multiplication lets us see every factor.
x = x
Factorise the second bottom: x − 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x − 2 is a single bracket already. It cannot be factorised any further.
x − 2 = (x − 2)
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
(x − 2)(x + 2)(x + 2) × x(x − 2)
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
(x − 2)(x + 2) × x(x + 2) × (x − 2)
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x − 2), (x + 2).
~~(x − 2)~~x~~(x + 2)(x − 2)(x + 2)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
x
State the result. The last line must match the result we were asked to show.
x² − 4x + 2 × xx − 2 = x ✓

Check. Try x = 3 (it does not make any bottom zero). The original expression gives 3. Our answer gives 3. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

Done

RChallenge. Simplify x² − 5x + 6x² − 4 ÷ x − 3x + 2

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal, which is the fraction turned upside down. Here is why it works with numbers: 1/2 ÷ 1/4 asks "how many quarters fit into a half?" The answer is 2, and 1/2 × 4/1 = 2. So: keep the first fraction, change ÷ to ×, and flip the second fraction.
x² − 5x + 6x² − 4 ÷ x − 3x + 2 = x² − 5x + 6x² − 4 × x + 2x − 3
Factorise the first top: x² − 5x + 6 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -5 and c = 6.
We need two numbers that multiply to 6 (the last number) and add to -5 (the middle number). List the pairs that multiply to 6:
1 × 6 = 6, add: 1 + 6 = 7 ✗
-1 × -6 = 6, add: -1 + -6 = -7 ✗
2 × 3 = 6, add: 2 + 3 = 5 ✗
-2 × -3 = 6, add: -2 + -3 = -5 ✓ this pair works
Put the pair into two brackets: (x − 2)(x − 3).
Check by expanding: x² − 3x − 2x + 6 = x² − 5x + 6. ✓
x² − 5x + 6 = (x − 3)(x − 2)
Factorise the first bottom: x² − 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. x² = (x)², so a = x. And 4 = 2², so b = 2.
Put them into the rule: (x − 2)(x + 2).
Check by expanding: (x − 2)(x + 2) = x² + 2x − 2x − 4 = x² − 4. ✓ (the middle terms cancel)
x² − 4 = (x − 2)(x + 2)
Factorise the second top: x + 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x + 2 is a single bracket already. It cannot be factorised any further.
x + 2 = (x + 2)
Factorise the second bottom: x − 3 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
x − 3 is a single bracket already. It cannot be factorised any further.
x − 3 = (x − 3)
Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
(x − 3)(x − 2)(x − 2)(x + 2) × (x + 2)(x − 3)
Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
(x − 3)(x − 2) × (x + 2)(x − 2)(x + 2) × (x − 3)
Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
Matching factors: (x − 3), (x − 2), (x + 2).
~~(x − 3)(x − 2)(x + 2)(x − 3)(x − 2)(x + 2)~~
(the crossed-out factors cancel)
Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
1

Check. Try x = 4 (it does not make any bottom zero). The original expression gives 1. Our answer gives 1. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

Done

3.5

Algebraic fractions and equations

Edexcel 1MA1: A4, A17
Grade 6–8

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Algebraic fractions and equationsComing soon

To add or subtract algebraic fractions, use a common denominator exactly as with number fractions. To solve an equation with fractions, multiply every term by the common denominator to clear them.

Key ideas

Find a common denominator (often the product of the denominators, or the LCM of them).
Rewrite each fraction with that denominator, then combine the numerators. Expand and simplify the numerator; leave the denominator factorised.
To solve an equation, multiply every term by the common denominator.
Check solutions by substituting back, especially when x is in a denominator (the denominator must not be 0).

Worked example 3.5.1

Write 2x + 1 + 3x − 2 as a single fraction.

Key words. The denominator is the bottom of a fraction and the numerator is the top. A common denominator is a bottom that all the fractions share. This works exactly like 1/2 + 1/3 = 3/6 + 2/6 = 5/6, except that the bottoms contain letters.

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
2x + 1 + 3x − 2
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
x + 1 = (x + 1)
x − 2 = (x − 2)
Common denominator = (x − 2)(x + 1)
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
2x + 1: to turn (x + 1) into (x − 2)(x + 1) we multiply the bottom by (x − 2). We must do the same to the top, so the value does not change:
2(x − 2)(x − 2)(x + 1)
3x − 2: to turn (x − 2) into (x − 2)(x + 1) we multiply the bottom by (x + 1). We must do the same to the top, so the value does not change:
3(x + 1)(x − 2)(x + 1)
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
2(x − 2) + 3(x + 1)(x − 2)(x + 1)
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 2(x − 2):
2 × x = 2x
2 × (−2) = −4
so 2(x − 2) = 2x − 4

Expand 3(x + 1):
3 × x = 3x
3 × 1 = 3
so 3(x + 1) = 3x + 3
Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
Top = 2x − 4 + 3x + 3
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 5x − 1
Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
5x − 1(x − 2)(x + 1)
The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 3 (it does not make any bottom zero). The original gives 7/2. Our answer gives 7/2. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Worked example 3.5.2

Write x + 23 − x − 14 as a single fraction.

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
x + 23 − x − 14
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
3 = 3
4 = 4
Common denominator = 12
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
x + 23: to turn 3 into 12 we multiply the bottom by 4. We must do the same to the top, so the value does not change:
4(x + 2)12
x − 14: to turn 4 into 12 we multiply the bottom by 3. We must do the same to the top, so the value does not change:
3(x − 1)12
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
4(x + 2) − 3(x − 1)12
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 4(x + 2):
4 × x = 4x
4 × 2 = 8
so 4(x + 2) = 4x + 8

Expand 3(x − 1):
3 × x = 3x
3 × (−1) = −3
so 3(x − 1) = 3x − 3
Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it. A plus sign changes nothing.
Top = (4x + 8) − (3x − 3)
Top = 4x + 8 − 3x + 3
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = x + 11
Check whether the answer can be simplified. The top might factorise and share a factor with the bottom.
x + 11 is a single bracket already. It cannot be factorised any further.
(x + 11)12 = 1/12(x + 11)

Check. Try x = 2 (it does not make any bottom zero). The original gives 13/12. Our answer gives 13/12. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Worked example 3.5.3

Solve 2x + 13 − x − 12 = 1

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
2x + 13 − x − 12 = 1
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
3, 2
Common denominator = 6
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
6 × 2x + 13 = 2(2x + 1) (the bottom 3 goes into 6 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
6 × x − 12 = 3(x − 1) (the bottom 2 goes into 6 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
6 × 1 = 6 (a plain number or term with no bottom is simply multiplied by 6)
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
2(2x + 1) − 3(x − 1) = 6
Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
Expand 2(2x + 1):
2 × 2x = 4x
2 × 1 = 2
so it is 4x + 2

Expand 3(x − 1):
3 × x = 3x
3 × (−1) = −3
so it is 3x − 3
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
4x + 2 − 3x + 3 = 6
x + 5 = 6
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: x + 5 = 6
Now get the numbers on the other side. Subtract 5 from both sides: x = 1
Read off the answer. x is already on its own.
x = 1

Check. Put x = 1 back into the original equation. Left side: 3/3 − 0/2 = 1. Right side: 1 = 1. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Worked example 3.5.4

Solve 5x + 1 = 2x − 2

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
5x + 1 = 2x − 2
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
x + 1, x − 2
Common denominator = (x − 2)(x + 1)
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
(x² − x − 2) × 5x + 1 = 5(x − 2) (the bottom x + 1 goes into (x − 2)(x + 1) exactly (x − 2) times, so the bottom disappears and the top is multiplied by (x − 2))
(x² − x − 2) × 2x − 2 = 2(x + 1) (the bottom x − 2 goes into (x − 2)(x + 1) exactly (x + 1) times, so the bottom disappears and the top is multiplied by (x + 1))
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
5(x − 2) = 2(x + 1)
Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
Expand 5(x − 2):
5 × x = 5x
5 × (−2) = −10
so it is 5x − 10

Expand 2(x + 1):
2 × x = 2x
2 × 1 = 2
so it is 2x + 2
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
5x − 10 = 2x + 2
5x − 10 = 2x + 2
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: 5x − 10 = 2x + 2
The x terms should be on one side. Subtract 2x from both sides: 3x − 10 = 2
Now get the numbers on the other side. Add 10 to both sides: 3x = 12
Divide both sides by the number in front of x (which is 3). 3x means 3 multiplied by x. The inverse of multiplying by 3 is dividing by 3. This leaves x on its own.
x = 12 ÷ 3 = 4

Check. Put x = 4 back into the original equation. Left side: 5/5 = 1. Right side: 2/2 = 1. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Worked example 3.5.5

Solve x2 + x3 = 5

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
x2 + x3 = 5
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
2, 3
Common denominator = 6
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
6 × x2 = 3x (the bottom 2 goes into 6 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
6 × x3 = 2x (the bottom 3 goes into 6 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
6 × 5 = 30 (a plain number or term with no bottom is simply multiplied by 6)
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
3x + 2x = 30
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
3x + 2x = 30
5x = 30
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: 5x = 30
Divide both sides by the number in front of x (which is 5). 5x means 5 multiplied by x. The inverse of multiplying by 5 is dividing by 5. This leaves x on its own.
x = 30 ÷ 5 = 6

Check. Put x = 6 back into the original equation. Left side: 6/2 + 6/3 = 5. Right side: 5 = 5. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Exercise 3E Algebraic fractions and equations

Non-calculator

FWrite each as a single fraction in its simplest form.

(a)x3 + x4

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
x3 + x4
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
3 = 3
4 = 4
Common denominator = 12
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
x3: to turn 3 into 12 we multiply the bottom by 4. We must do the same to the top, so the value does not change:
4x12
x4: to turn 4 into 12 we multiply the bottom by 3. We must do the same to the top, so the value does not change:
3x12
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
4x + 3x12
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 4x:
x × 4 = 4x
so 4x = 4x

Expand 3x:
x × 3 = 3x
so 3x = 3x
Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
Top = 4x + 3x
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 7x
Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
7x12
The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 2 (it does not make any bottom zero). The original gives 7/6. Our answer gives 7/6. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

(b)2x5 − x10

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
2x5 − x10
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
5 = 5
10 = 10
Common denominator = 10
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
2x5: to turn 5 into 10 we multiply the bottom by 2. We must do the same to the top, so the value does not change:
22x10
x10: the bottom is already 10, so leave it.
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
22x − x10
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 22x:
2x × 2 = 4x
so 22x = 4x
Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
Top = 4x − x
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 3x
Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
3x10
The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 2 (it does not make any bottom zero). The original gives 3/5. Our answer gives 3/5. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
3x + 2x²
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
x = x
x² = x²
Common denominator = x²
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
3x: to turn x into x² we multiply the bottom by x. We must do the same to the top, so the value does not change:
3xx²
2x²: the bottom is already x², so leave it.
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
3x + 2x²
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 3x:
3 × x = 3x
so 3x = 3x
Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
Top = 3x + 2
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 3x + 2
Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
3x + 2x²
The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 2 (it does not make any bottom zero). The original gives 2. Our answer gives 2. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

(d)x + 12 + x − 13

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
x + 12 + x − 13
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
2 = 2
3 = 3
Common denominator = 6
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
x + 12: to turn 2 into 6 we multiply the bottom by 3. We must do the same to the top, so the value does not change:
3(x + 1)6
x − 13: to turn 3 into 6 we multiply the bottom by 2. We must do the same to the top, so the value does not change:
2(x − 1)6
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
3(x + 1) + 2(x − 1)6
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 3(x + 1):
3 × x = 3x
3 × 1 = 3
so 3(x + 1) = 3x + 3

Expand 2(x − 1):
2 × x = 2x
2 × (−1) = −2
so 2(x − 1) = 2x − 2
Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
Top = 3x + 3 + 2x − 2
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 5x + 1
Check whether the answer can be simplified. The top might factorise and share a factor with the bottom.
5x + 1 = (5x + 1)
(5x + 1)6 = 1/6(5x + 1)

Check. Try x = 2 (it does not make any bottom zero). The original gives 11/6. Our answer gives 11/6. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

(e)2x − 14 − x + 26

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
2x − 14 − x + 26
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
4 = 4
6 = 6
Common denominator = 12
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
2x − 14: to turn 4 into 12 we multiply the bottom by 3. We must do the same to the top, so the value does not change:
3(2x − 1)12
x + 26: to turn 6 into 12 we multiply the bottom by 2. We must do the same to the top, so the value does not change:
2(x + 2)12
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
3(2x − 1) − 2(x + 2)12
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 3(2x − 1):
3 × 2x = 6x
3 × (−1) = −3
so 3(2x − 1) = 6x − 3

Expand 2(x + 2):
2 × x = 2x
2 × 2 = 4
so 2(x + 2) = 2x + 4
Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it. A plus sign changes nothing.
Top = (6x − 3) − (2x + 4)
Top = 6x − 3 − 2x − 4
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 4x − 7
Check whether the answer can be simplified. The top might factorise and share a factor with the bottom.
4x − 7 = (4x − 7)
(4x − 7)12 = 1/12(4x − 7)

Check. Try x = 2 (it does not make any bottom zero). The original gives 1/12. Our answer gives 1/12. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Done

FWrite each as a single fraction. Leave the denominator factorised.

(a)1x + 1 + 1x + 2

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
1x + 1 + 1x + 2
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
x + 1 = (x + 1)
x + 2 = (x + 2)
Common denominator = (x + 1)(x + 2)
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
1x + 1: to turn (x + 1) into (x + 1)(x + 2) we multiply the bottom by (x + 2). We must do the same to the top, so the value does not change:
(x + 2)(x + 1)(x + 2)
1x + 2: to turn (x + 2) into (x + 1)(x + 2) we multiply the bottom by (x + 1). We must do the same to the top, so the value does not change:
(x + 1)(x + 1)(x + 2)
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
(x + 2) + (x + 1)(x + 1)(x + 2)
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand (x + 2):
1 × x = x
1 × 2 = 2
so (x + 2) = x + 2

Expand (x + 1):
1 × x = x
1 × 1 = 1
so (x + 1) = x + 1
Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
Top = x + 2 + x + 1
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 2x + 3
Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
2x + 3(x + 1)(x + 2)
The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 2 (it does not make any bottom zero). The original gives 7/12. Our answer gives 7/12. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

(b)2x − 1 − 3x + 2

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
2x − 1 − 3x + 2
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
x − 1 = (x − 1)
x + 2 = (x + 2)
Common denominator = (x − 1)(x + 2)
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
2x − 1: to turn (x − 1) into (x − 1)(x + 2) we multiply the bottom by (x + 2). We must do the same to the top, so the value does not change:
2(x + 2)(x − 1)(x + 2)
3x + 2: to turn (x + 2) into (x − 1)(x + 2) we multiply the bottom by (x − 1). We must do the same to the top, so the value does not change:
3(x − 1)(x − 1)(x + 2)
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
2(x + 2) − 3(x − 1)(x − 1)(x + 2)
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 2(x + 2):
2 × x = 2x
2 × 2 = 4
so 2(x + 2) = 2x + 4

Expand 3(x − 1):
3 × x = 3x
3 × (−1) = −3
so 3(x − 1) = 3x − 3
Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it. A plus sign changes nothing.
Top = (2x + 4) − (3x − 3)
Top = 2x + 4 − 3x + 3
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = −x + 7
Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
−x + 7(x − 1)(x + 2)
The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 2 (it does not make any bottom zero). The original gives 5/4. Our answer gives 5/4. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
3x + 2 + 4x − 3
Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
x + 2 = (x + 2)
x − 3 = (x − 3)
Common denominator = (x − 3)(x + 2)
Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
3x + 2: to turn (x + 2) into (x − 3)(x + 2) we multiply the bottom by (x − 3). We must do the same to the top, so the value does not change:
3(x − 3)(x − 3)(x + 2)
4x − 3: to turn (x − 3) into (x − 3)(x + 2) we multiply the bottom by (x + 2). We must do the same to the top, so the value does not change:
4(x + 2)(x − 3)(x + 2)
Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
3(x − 3) + 4(x + 2)(x − 3)(x + 2)
Expand each product on the top. Multiply every term in one bracket by every term in the other.
Expand 3(x − 3):
3 × x = 3x
3 × (−3) = −9
so 3(x − 3) = 3x − 9

Expand 4(x + 2):
4 × x = 4x
4 × 2 = 8
so 4(x + 2) = 4x + 8
Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
Top = 3x − 9 + 4x + 8
Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
Top = 7x − 1
Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
7x − 1(x − 3)(x + 2)
The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 2 (it does not make any bottom zero). The original gives -13/4. Our answer gives -13/4. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Done

FSolve.

(a)x3 + x6 = 4

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
x3 + x6 = 4
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
3, 6
Common denominator = 6
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
6 × x3 = 2x (the bottom 3 goes into 6 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
6 × x6 = x (the bottom 6 goes into 6 exactly 1 time, so the bottom disappears and the top is multiplied by 1)
6 × 4 = 24 (a plain number or term with no bottom is simply multiplied by 6)
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
2x + x = 24
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
2x + x = 24
3x = 24
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: 3x = 24
Divide both sides by the number in front of x (which is 3). 3x means 3 multiplied by x. The inverse of multiplying by 3 is dividing by 3. This leaves x on its own.
x = 24 ÷ 3 = 8

Check. Put x = 8 back into the original equation. Left side: 8/3 + 8/6 = 4. Right side: 4 = 4. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

(b)2x + 13 − x − 12 = 1

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
2x + 13 − x − 12 = 1
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
3, 2
Common denominator = 6
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
6 × 2x + 13 = 2(2x + 1) (the bottom 3 goes into 6 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
6 × x − 12 = 3(x − 1) (the bottom 2 goes into 6 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
6 × 1 = 6 (a plain number or term with no bottom is simply multiplied by 6)
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
2(2x + 1) − 3(x − 1) = 6
Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
Expand 2(2x + 1):
2 × 2x = 4x
2 × 1 = 2
so it is 4x + 2

Expand 3(x − 1):
3 × x = 3x
3 × (−1) = −3
so it is 3x − 3
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
4x + 2 − 3x + 3 = 6
x + 5 = 6
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: x + 5 = 6
Now get the numbers on the other side. Subtract 5 from both sides: x = 1
Read off the answer. x is already on its own.
x = 1

Check. Put x = 1 back into the original equation. Left side: 3/3 − 0/2 = 1. Right side: 1 = 1. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

(c)x + 32 = 2x − 13

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
x + 32 = 2x − 13
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
2, 3
Common denominator = 6
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
6 × x + 32 = 3(x + 3) (the bottom 2 goes into 6 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
6 × 2x − 13 = 2(2x − 1) (the bottom 3 goes into 6 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
3(x + 3) = 2(2x − 1)
Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
Expand 3(x + 3):
3 × x = 3x
3 × 3 = 9
so it is 3x + 9

Expand 2(2x − 1):
2 × 2x = 4x
2 × (−1) = −2
so it is 4x − 2
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
3x + 9 = 4x − 2
3x + 9 = 4x − 2
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: 3x + 9 = 4x − 2
Subtract 3x from both sides (we move the smaller x term, to keep the numbers positive): 9 = x − 2
Now add 2 to both sides: 11 = x
Read off the answer. x is already on its own.
x = 11

Check. Put x = 11 back into the original equation. Left side: 14/2 = 7. Right side: 21/3 = 7. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

(d)5x + 1 = 2x − 2

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
5x + 1 = 2x − 2
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
x + 1, x − 2
Common denominator = (x − 2)(x + 1)
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
(x² − x − 2) × 5x + 1 = 5(x − 2) (the bottom x + 1 goes into (x − 2)(x + 1) exactly (x − 2) times, so the bottom disappears and the top is multiplied by (x − 2))
(x² − x − 2) × 2x − 2 = 2(x + 1) (the bottom x − 2 goes into (x − 2)(x + 1) exactly (x + 1) times, so the bottom disappears and the top is multiplied by (x + 1))
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
5(x − 2) = 2(x + 1)
Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
Expand 5(x − 2):
5 × x = 5x
5 × (−2) = −10
so it is 5x − 10

Expand 2(x + 1):
2 × x = 2x
2 × 1 = 2
so it is 2x + 2
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
5x − 10 = 2x + 2
5x − 10 = 2x + 2
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: 5x − 10 = 2x + 2
The x terms should be on one side. Subtract 2x from both sides: 3x − 10 = 2
Now get the numbers on the other side. Add 10 to both sides: 3x = 12
Divide both sides by the number in front of x (which is 3). 3x means 3 multiplied by x. The inverse of multiplying by 3 is dividing by 3. This leaves x on its own.
x = 12 ÷ 3 = 4

Check. Put x = 4 back into the original equation. Left side: 5/5 = 1. Right side: 2/2 = 1. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

(e)3x + 12 = 2

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
3x + 12 = 2
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
x, 2
Common denominator = 2x
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
2x × 3x = 6 (the bottom x goes into 2x exactly 2 times, so the bottom disappears and the top is multiplied by 2)
2x × 12 = x (the bottom 2 goes into 2x exactly x times, so the bottom disappears and the top is multiplied by x)
2x × 2 = 22x (a plain number or term with no bottom is simply multiplied by 2x)
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
6 + x = 22x
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
6 + x = 4x
x + 6 = 4x
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: x + 6 = 4x
Subtract x from both sides (we move the smaller x term, to keep the numbers positive): 6 = 3x
Divide both sides by the number in front of x (which is 3). 3x means 3 multiplied by x. The inverse of multiplying by 3 is dividing by 3. This leaves x on its own.
x = 6 ÷ 3 = 2

Check. Put x = 2 back into the original equation. Left side: 3/2 + 1/2 = 2. Right side: 2 = 2. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

(f)x + 4x − 1 = 3

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
x + 4x − 1 = 3
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
x − 1
Common denominator = (x − 1)
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
(x − 1) × x + 4x − 1 = (x + 4) (the bottom x − 1 goes into (x − 1) exactly 1 time, so the bottom disappears and the top is multiplied by 1)
(x − 1) × 3 = 3(x − 1) (a plain number or term with no bottom is simply multiplied by (x − 1))
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
(x + 4) = 3(x − 1)
Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
Expand 3(x − 1):
3 × x = 3x
3 × (−1) = −3
so it is 3x − 3
Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
x + 4 = 3x − 3
x + 4 = 3x − 3
Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: x + 4 = 3x − 3
Subtract x from both sides (we move the smaller x term, to keep the numbers positive): 4 = 2x − 3
Now add 3 to both sides: 7 = 2x
Divide both sides by the number in front of x (which is 2). 2x means 2 multiplied by x. The inverse of multiplying by 2 is dividing by 2. This leaves x on its own.
x = 72 = 7/2

Check. Put x = 7/2 back into the original equation. Left side: 15/2/5/2 = 3. Right side: 3 = 3. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Done

PSOne third of a number plus one fifth of the same number is 16. Find the number.

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

Turn the words into an equation. Let the number be n. One third of n is n over 3, and one fifth of n is n over 5. “Is 16” means equals 16.
n3 + n5 = 16
Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
n3 + n5 = 16
Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
3, 5
Common denominator = 15
Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
15 × n3 = 5n (the bottom 3 goes into 15 exactly 5 times, so the bottom disappears and the top is multiplied by 5)
15 × n5 = 3n (the bottom 5 goes into 15 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
15 × 16 = 240 (a plain number or term with no bottom is simply multiplied by 15)
Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
5n + 3n = 240
Collect like terms on each side. Add or subtract the n terms together, and the plain numbers together, on each side separately.
5n + 3n = 240
8n = 240
Get the n terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
Start: 8n = 240
Divide both sides by the number in front of n (which is 8). 8n means 8 multiplied by n. The inverse of multiplying by 8 is dividing by 8. This leaves n on its own.
n = 240 ÷ 8 = 30
Answer the question in words. The question asks for the number.
The number is 30.

Check. Put n = 30 back into the original equation. Left side: 30/3 + 30/5 = 16. Right side: 16 = 16. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Done

PSThe perimeter of a rectangle is 30 cm. The length is (x + 3) cm and the width is (x − 1) cm. Find x.

Done

RZak writes 1x + 1y = 2x + y. Show that this is wrong by choosing values for x and y.

Done

RChallenge. Show that 1n − 1n + 1 = 1n(n + 1).

Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.

Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.
Start from the left side: 1n − 1n + 1
Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.
Common denominator = n(n + 1)
Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.
1n = 1 × (n + 1)n(n + 1)
Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.
1n + 1 = 1 × nn(n + 1)
Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.
1 × (n + 1) − (1 × n)n(n + 1)
Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.
Expand 1 × (n + 1):
1 × n = n
1 × 1 = 1
so 1 × (n + 1) = n + 1

1 × n = n
Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.
Top = (n + 1) − n
Top = n + 1 − n
Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.
Top = 1
Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.
1n(n + 1)
State the result. The last line matches the result we were asked to show.
1n − 1n + 1 = 1n(n + 1) ✓

Check. Try n = 2. The original gives 1/6. Our result gives 1/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓

Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.

Done

3.6

Reasoning and proof with fractions

Edexcel 1MA1: N10, A4, A5
Grade 7–9

Printable: Worksheet (PDF) · Answers (PDF) · Open this lesson on its own page

Video lesson: Reasoning and proof with fractionsComing soon

The hardest exam questions ask you to show or prove a result. You must give a chain of correct steps that ends exactly at the stated result. Use algebra so that the proof works for every value, not only one example.

Key ideas

“Show that” means every step must appear. Start from one side and reach the other.
An example does not prove a general result. To prove something for all n, use algebra.
To disprove a statement, one counter-example is enough.

Worked example 3.6.1

Show that 1n − 1n + 1 = 1n(n + 1)

Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.

Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.
Start from the left side: 1n − 1n + 1
Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.
Common denominator = n(n + 1)
Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.
1n = 1 × (n + 1)n(n + 1)
Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.
1n + 1 = 1 × nn(n + 1)
Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.
1 × (n + 1) − (1 × n)n(n + 1)
Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.
Expand 1 × (n + 1):
1 × n = n
1 × 1 = 1
so 1 × (n + 1) = n + 1

1 × n = n
Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.
Top = (n + 1) − n
Top = n + 1 − n
Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.
Top = 1
Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.
1n(n + 1)
State the result. The last line matches the result we were asked to show.
1n − 1n + 1 = 1n(n + 1) ✓

Check. Try n = 2. The original gives 1/6. Our result gives 1/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓

Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.

Worked example 3.6.2

Use the result above to show that 11×2 + 12×3 + 13×4 + … + 19×10 = 910

Method. Rewrite every term as a difference using the result above. Then most terms cancel in pairs (a telescoping sum), leaving only the first and the last.

Recall the result. We have just shown that one over n(n + 1) equals one over n minus one over (n + 1).
1n(n + 1) = 1n − 1n + 1
Rewrite the first terms. Put n = 1, 2, 3, … into the result.
11×2 = 1 − 12
12×3 = 12 − 13
13×4 = 13 − 14
Rewrite the last term. Put n = 9.
19×10 = 19 − 110
Add everything. Write the whole sum with the differences in brackets.
(1 − 12) + (12 − 13) + (13 − 14) + … + (19 − 110)
Cancel the pairs. The −12 cancels with the next +12, the −13 cancels with the next +13, and so on all the way along.
Only 1 and −110 are left
Work out what is left. Write 1 as tenths and subtract.
1 − 110 = 1010 − 110 = 910

Check. Add the first few terms to see the pattern: 12 = 12; + 16 = 23; + 112 = 34; + 120 = 45. With n terms the total is n over (n + 1), so with 9 terms it is 910. ✓

Watch out. Write enough terms to show which ones cancel. Do not just state the answer.

Worked example 3.6.3

Prove that nn + 1 is always greater than n − 1n for positive integers n.

Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.

Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.
Subtract the smaller from the bigger: nn + 1 − n − 1n
Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n + 1 and n have no common factor (they are consecutive numbers), so the lowest common denominator is their product.
Common denominator = n(n + 1)
Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by n, the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.
nn + 1 = n × nn(n + 1)
Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1).
n − 1n = (n − 1) × (n + 1)n(n + 1)
Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.
n × n − ((n − 1) × (n + 1))n(n + 1)
Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.
n × n = n²

Expand (n − 1) × (n + 1):
First: n × n = n²
Outer: n × 1 = n
Inner: (−1) × n = −n
Last: (−1) × 1 = −1
so (n − 1) × (n + 1) = n² − 1
Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.
Top = n² − (n² − 1)
Top = n² − n² + 1
Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.
Top = 1
Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.
1n(n + 1)
Decide the sign. The difference is 1 divided by n(n + 1). When n is a positive integer, both n and n + 1 are positive, so the bottom is positive, and the top is 1. A positive number divided by a positive number is positive.
1n(n + 1) > 0 for every positive integer n
State the result. The last line matches the result we were asked to show.
nn + 1 > n − 1n for every positive integer n ✓

Check. Try n = 2. The original gives 1/6. Our result gives 1/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓

Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.

Worked example 3.6.4

Alex says “A fraction with a bigger denominator is always smaller.” Give a counter-example to show Alex is wrong.

Method. A counter-example is one example that makes the statement false. To disprove a statement, one counter-example is enough.

Say what the statement claims. Alex says that if fraction B has a bigger denominator than fraction A, then B must be smaller than A.
Claim: bigger denominator ⇒ smaller fraction
Choose two fractions to test the claim. The claim is true when the numerators are the same. To break it, make the numerator of the fraction with the bigger denominator much bigger.
Try 14 and 58: 8 is bigger than 4
Give both fractions the same denominator. The LCM of 4 and 8 is 8.
14 = 28
Compare them. Same denominator, so compare the numerators.
58 > 28, so 58 > 14
Write the conclusion. The fraction with the bigger denominator (58) is the bigger fraction, which breaks Alex's rule.
Alex is wrong.

Check. As decimals: 58 = 0.625 and 14 = 0.25, so 58 is bigger even though 8 > 4. ✓

Watch out. A counter-example must be a specific case that you have checked. Saying “it is not always true” without an example earns no marks.

Exercise 3F Reasoning and proof

Non-calculator

RShow that 1n + 1n + 1 = 2n + 1n(n + 1).

Method. Use a common denominator to add the fractions, expand and collect the top, and compare with the result.

Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.
Start from the left side: 1n + 1n + 1
Find a common denominator. We can only add fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.
Common denominator = n(n + 1)
Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.
1n = 1 × (n + 1)n(n + 1)
Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.
1n + 1 = 1 × nn(n + 1)
Combine the tops over the common denominator. Now the pieces are the same size, so we add the tops and keep the bottom.
1 × (n + 1) + 1 × nn(n + 1)
Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.
Expand 1 × (n + 1):
1 × n = n
1 × 1 = 1
so 1 × (n + 1) = n + 1

1 × n = n
Write the top without brackets. Adding changes no signs.
Top = n + 1 + n
Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.
Top = 2n + 1
Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.
2n + 1n(n + 1)
State the result. The last line matches the result we were asked to show.
1n + 1n + 1 = 2n + 1n(n + 1) ✓

Check. Try n = 2. The original gives 5/6. Our result gives 5/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓

Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.

Done

RWork out 11×2 + 12×3 + … + 119×20.

Done

RProve that 0.18 = 211.

Done

RShow that 35 lies between 12 and 23. Explain how adding the numerators and adding the denominators of 12 and 23 gives it.

Done

RShow that n + 1n − nn + 1 = 2n + 1n(n + 1).

Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.

Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.
Start from the left side: n + 1n − nn + 1
Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.
Common denominator = n(n + 1)
Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.
n + 1n = (n + 1) × (n + 1)n(n + 1)
Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.
nn + 1 = n × nn(n + 1)
Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.
(n + 1) × (n + 1) − (n × n)n(n + 1)
Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.
Expand (n + 1) × (n + 1):
First: n × n = n²
Outer: n × 1 = n
Inner: 1 × n = n
Last: 1 × 1 = 1
so (n + 1) × (n + 1) = n² + 2n + 1

n × n = n²
Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.
Top = (n² + 2n + 1) − n²
Top = n² + 2n + 1 − n²
Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.
Top = 2n + 1
Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.
2n + 1n(n + 1)
State the result. The last line matches the result we were asked to show.
n + 1n − nn + 1 = 2n + 1n(n + 1) ✓

Check. Try n = 2. The original gives 5/6. Our result gives 5/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓

Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.

Done

RBen says “The sum of two proper fractions is always a proper fraction.” Give a counter-example.

Done

RWrite 1 as the sum of three different unit fractions.

Done

RChallenge. Find the positive whole numbers a and b with a < b such that 1a + 1b = 12.

Done

Chapter summary

Skill	Method
Brackets and mixed numbers	BIDMAS; mixed → improper first; simplify at the end
Fractional change	Increase by a/b: × (1 + a/b). Decrease: × (1 − a/b). Multiply successive multipliers
Reverse problems	final ÷ multiplier = original
Terminating test	Simplest form; denominator only has prime factors 2 and 5
Recurring → fraction	x = decimal; multiply by 10, 100… to line up; subtract; solve
Simplify algebraic fractions	Factorise top and bottom, cancel common factors (never terms)
Multiply / divide algebraic	Factorise, flip the second for division, cancel
Add / subtract algebraic	Common denominator, combine numerators, expand and simplify
Equations with fractions	Multiply every term by the common denominator
Proof	Use algebra; each step shown; end exactly at the stated result

Can you…?

I can handle brackets, mixed numbers and negative fractions.
I can solve multi-step and reverse fraction problems.
I can decide if a fraction terminates, and convert recurring decimals to fractions.
I can simplify, multiply and divide algebraic fractions.
I can add and subtract algebraic fractions.
I can solve equations containing fractions.
I can write short algebraic proofs involving fractions.

Key terms

Algebraic fraction: A fraction with algebra in the numerator or denominator.
Common factor: A factor shared by two or more terms or numbers.
Counter-example: A single example that shows a statement is false.
Multiplier: The number you multiply by to apply a percentage or fractional change.
Recurring decimal: A decimal with a digit or block of digits that repeats forever.
Telescoping sum: A sum in which most terms cancel in pairs.
Terminating decimal: A decimal that ends.

Total 49 marks. Allow about 49 minutes. Non-calculator.

GCSE Maths Fractions (Higher): worked examples and questions with answers

By the end of this chapter you will be able to

You should already know

Where this leads

Exercise P Do you remember?

Fraction arithmetic with confidence

Key ideas

Watch out

Exercise 3A Fraction arithmetic

Multi-step and reverse fraction problems

Key ideas

Watch out

Exercise 3B Multi-step and reverse problems

Recurring decimals and fractions

Key ideas

Exam tip

Exercise 3C Recurring decimals

Simplifying algebraic fractions

Key ideas

Watch out

Exercise 3D Simplifying, multiplying and dividing algebraic fractions

Algebraic fractions and equations

Key ideas

Exercise 3E Algebraic fractions and equations

Reasoning and proof with fractions

Key ideas

Exercise 3F Reasoning and proof

Chapter summary

Can you…?

Key terms

Exercise 3G Check-up (non-calculator)

Exercise 3H Exam-style questions