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GCSE Maths · EdexcelChapter 3 · Fractions
GCSE Maths · Fractions · Higher · Lesson 3.5

GCSE Maths: Algebraic fractions and equations (Higher)

Part of the GCSE Maths fractions chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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3.5

Algebraic fractions and equations

Edexcel 1MA1: A4, A17
Grade 6–8

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.

  1. Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
    2x + 1 + 3x − 2
  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)
  3. 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)
  4. 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)
  5. 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
  6. 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
  7. 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
  8. 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.

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.

  1. Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
    x + 23 − x − 14
  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.
    3 = 3
    4 = 4
    Common denominator = 12
  3. 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
  4. 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
  5. 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
  6. 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
  7. 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
  8. 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.

  1. 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
  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.
    3, 2
    Common denominator = 6
  3. 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)
  4. 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
  5. 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
  6. 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
  7. 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
  8. 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.

  1. 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
  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)
  3. 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))
  4. 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)
  5. 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
  6. 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
  7. 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
  8. 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.

  1. Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
    x2 + x3 = 5
  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.
    2, 3
    Common denominator = 6
  3. 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)
  4. Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
    3x + 2x = 30
  5. 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
  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: 5x = 30
  7. 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
1
FWrite each as a single fraction in its simplest form.
(a)x3 + x4
(b)2x5 − x10
(c)3x + 2x²
(d)x + 12 + x − 13
(e)2x − 14 − x + 26
2
FWrite each as a single fraction. Leave the denominator factorised.
(a)1x + 1 + 1x + 2
(b)2x − 1 − 3x + 2
(c)3x + 2 + 4x − 3
3
FSolve.
(a)x3 + x6 = 4
(b)2x + 13 − x − 12 = 1
(c)x + 32 = 2x − 13
(d)5x + 1 = 2x − 2
(e)3x + 12 = 2
(f)x + 4x − 1 = 3
4
PSOne third of a number plus one fifth of the same number is 16. Find the number.
5
PSThe perimeter of a rectangle is 30 cm. The length is (x + 3) cm and the width is (x − 1) cm. Find x.
6
RZak writes 1x + 1y = 2x + y. Show that this is wrong by choosing values for x and y.
7
RChallenge. Show that 1n − 1n + 1 = 1n(n + 1).

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