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GCSE Maths · EdexcelChapter 9 · Linear equations
GCSE Maths · Linear Equations · Higher · Lesson 9.6

GCSE Maths: Equations with fractions (Higher)

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

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9.6

Equations with fractions

Edexcel 1MA1: A17
Grade 4-7

In Chapter 3 you added and subtracted fractions by changing them to the same bottom number (the common denominator). We use the same idea to solve equations that contain fractions, but we use it to remove the fractions altogether.

Consider x + 32 = 5. The whole of the top, x + 3, is divided by 2. To undo “divide by 2” we multiply both sides by 2: x + 3 = 10, so x = 7. On the left, 2 × x + 32 = x + 3, because the 2 on the top and the 2 on the bottom cancel. Note that the top is a bracket in disguise: the fraction bar acts like a bracket around x + 3.

When there is more than one fraction, for example x2 + x3 = 10, find the lowest common multiple (LCM) of the bottom numbers. Here the LCM of 2 and 3 is 6. Multiply every term on both sides by 6:

6 × x2 = 3x    6 × x3 = 2x    6 × 10 = 60, so 3x + 2x = 60, which gives 5x = 60 and x = 12.

Why does this work? Multiplying both sides of an equation by the same number keeps it balanced. If we choose a number that every bottom divides into, each fraction turns into a whole expression. It is exactly like the way you made equivalent fractions in Chapter 3, only now the aim is to get rid of the bottoms.

The most common slips are (1) forgetting to multiply a term that is not a fraction, such as the 10 above, and (2) forgetting that a numerator with two terms needs a bracket when you multiply: 6 × x + 12 = 3(x + 1), not 3x + 1.

Key ideas

  • To solve an equation with fractions, multiply every term on both sides by the lowest common multiple of the bottom numbers.
  • This removes the fractions: after you multiply, the bottoms cancel.
  • If a numerator has two terms, use a bracket after multiplying, then expand it.
  • Multiply every term, including whole numbers. Then solve the ordinary equation that is left, and check in the original equation.
Worked example 9.6.1
Solve x + 32 = 5.

Key words. An equation says that two things are equal, for example 3x + 5 = 20. To solve it means to find the value of the letter that makes it true. The inverse operation is the operation that undoes another: adding and subtracting undo each other, and multiplying and dividing undo each other. An equation is like a pair of balanced scales: whatever you do to one side you must do to the other side. A common denominator is a bottom number that all the fractions can be changed to; multiplying every term by it removes the fractions.

Method. Solve the equation by using inverse operations (the opposite operation) on both sides, so that the equation stays balanced, until the letter x is on its own.

  1. Write down the equation. The two sides of an equation are equal, like the two sides of a balanced pair of scales. We want the value of x that keeps them balanced.
    x + 32 = 5
  2. Find the common denominator. To get rid of fractions we multiply by a number that every bottom divides into exactly: the lowest common multiple of the bottoms.
    2  →  the lowest common multiple is 2
  3. Multiply every term, on both sides, by 2. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers. This removes the fractions.
    2 × x + 32 = (x + 3)   (2 goes into 2 exactly 1 time, so the bottom disappears)
    2 × 5 = 10
    The new equation is (x + 3) = 10
  4. Expand the brackets. Multiply each term inside the bracket by the number outside it.
    (x + 3) has nothing in front of it except +1, so the brackets can simply be removed: x + 3

    The equation is now x + 3 = 10
  5. Subtract 3 from both sides. This undoes the +3, because +3 − 3 = 0, which leaves the x term on its own. We do it to both sides to keep the balance.
    x + 3 − 3 = 10 − 3
    x = 7
  6. Write the solution. x is now on its own, so the equation tells us its value.
    x = 7

Check. Substitute x = 7 into the original equation. Left side: 7 + 32 = 102 = 5. Right side: 5. Both sides equal 5, so the solution is correct. ✓

Watch out. Multiply every term by the common denominator, including the terms that are not fractions. Then check that no fraction is left.

Worked example 9.6.2
Solve 2x − 13 = 5.

Key words. An equation says that two things are equal, for example 3x + 5 = 20. To solve it means to find the value of the letter that makes it true. The inverse operation is the operation that undoes another: adding and subtracting undo each other, and multiplying and dividing undo each other. An equation is like a pair of balanced scales: whatever you do to one side you must do to the other side. A common denominator is a bottom number that all the fractions can be changed to; multiplying every term by it removes the fractions.

Method. Solve the equation by using inverse operations (the opposite operation) on both sides, so that the equation stays balanced, until the letter x is on its own.

  1. Write down the equation. The two sides of an equation are equal, like the two sides of a balanced pair of scales. We want the value of x that keeps them balanced.
    2x − 13 = 5
  2. Find the common denominator. To get rid of fractions we multiply by a number that every bottom divides into exactly: the lowest common multiple of the bottoms.
    3  →  the lowest common multiple is 3
  3. Multiply every term, on both sides, by 3. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers. This removes the fractions.
    3 × 2x − 13 = (2x − 1)   (3 goes into 3 exactly 1 time, so the bottom disappears)
    3 × 5 = 15
    The new equation is (2x − 1) = 15
  4. Expand the brackets. Multiply each term inside the bracket by the number outside it.
    (2x − 1) has nothing in front of it except +1, so the brackets can simply be removed: 2x − 1

    The equation is now 2x − 1 = 15
  5. Add 1 to both sides. This undoes the −1, because −1 + 1 = 0, which leaves the x term on its own. We do it to both sides to keep the balance.
    2x − 1 + 1 = 15 + 1
    2x = 16
  6. Divide both sides by 2. 2x means 2 multiplied by x. The inverse of multiplying by 2 is dividing by 2, and we do it to both sides to keep the balance. This leaves x on its own.
    2x2 = 162
    x = 162 = 8
  7. Write the solution. x is now on its own, so the equation tells us its value.
    x = 8

Check. Substitute x = 8 into the original equation. Left side: 2 × 8 − 13 = 153 = 5. Right side: 5. Both sides equal 5, so the solution is correct. ✓

Watch out. Multiply every term by the common denominator, including the terms that are not fractions. Then check that no fraction is left.

Worked example 9.6.3
Solve x2 + x3 = 10.

Key words. An equation says that two things are equal, for example 3x + 5 = 20. To solve it means to find the value of the letter that makes it true. The inverse operation is the operation that undoes another: adding and subtracting undo each other, and multiplying and dividing undo each other. An equation is like a pair of balanced scales: whatever you do to one side you must do to the other side. A common denominator is a bottom number that all the fractions can be changed to; multiplying every term by it removes the fractions.

Method. Solve the equation by using inverse operations (the opposite operation) on both sides, so that the equation stays balanced, until the letter x is on its own.

  1. Write down the equation. The two sides of an equation are equal, like the two sides of a balanced pair of scales. We want the value of x that keeps them balanced.
    x2 + x3 = 10
  2. Find the common denominator. To get rid of fractions we multiply by a number that every bottom divides into exactly: the lowest common multiple of the bottoms.
    2, 3  →  the lowest common multiple is 6
  3. Multiply every term, on both sides, by 6. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers. This removes the fractions.
    6 × x2 = 3x   (2 goes into 6 exactly 3 times, so the bottom disappears)
    6 × x3 = 2x   (3 goes into 6 exactly 2 times, so the bottom disappears)
    6 × 10 = 60
    The new equation is 3x + 2x = 60
  4. Collect like terms on each side. Like terms are terms with the same letter part: all the x terms go together, and all the plain numbers go together. Doing this makes each side as short as possible.
    On the left side: the x terms 3x + 2x make 5x.
    The equation is now 5x = 60
  5. Divide both sides by 5. 5x means 5 multiplied by x. The inverse of multiplying by 5 is dividing by 5, and we do it to both sides to keep the balance. This leaves x on its own.
    5x5 = 605
    x = 605 = 12
  6. Write the solution. x is now on its own, so the equation tells us its value.
    x = 12

Check. Substitute x = 12 into the original equation. Left side: 122 + 123 = 6 + 4 = 10. Right side: 10. Both sides equal 10, so the solution is correct. ✓

Watch out. Multiply every term by the common denominator, including the terms that are not fractions. Then check that no fraction is left.

Worked example 9.6.4
Solve x3 + 2 = x2.

Key words. An equation says that two things are equal, for example 3x + 5 = 20. To solve it means to find the value of the letter that makes it true. The inverse operation is the operation that undoes another: adding and subtracting undo each other, and multiplying and dividing undo each other. An equation is like a pair of balanced scales: whatever you do to one side you must do to the other side. A common denominator is a bottom number that all the fractions can be changed to; multiplying every term by it removes the fractions.

Method. Solve the equation by using inverse operations (the opposite operation) on both sides, so that the equation stays balanced, until the letter x is on its own.

  1. Write down the equation. The two sides of an equation are equal, like the two sides of a balanced pair of scales. We want the value of x that keeps them balanced.
    x3 + 2 = x2
  2. Find the common denominator. To get rid of fractions we multiply by a number that every bottom divides into exactly: the lowest common multiple of the bottoms.
    3, 2  →  the lowest common multiple is 6
  3. Multiply every term, on both sides, by 6. Doing the same to both sides keeps the equation balanced. Every term must be multiplied, including plain numbers. This removes the fractions.
    6 × x3 = 2x   (3 goes into 6 exactly 2 times, so the bottom disappears)
    6 × 2 = 12
    6 × x2 = 3x   (2 goes into 6 exactly 3 times, so the bottom disappears)
    The new equation is 2x + 12 = 3x
  4. Remove the x term from the left side: subtract 2x from both sides. We want all the x terms on one side. The left side has 2x; subtracting 2x removes it. We take away the smaller x term, so that the x term that is left is positive and easier to work with. We must do it to both sides to keep the balance.
    2x + 12 − 2x = 3x − 2x
    12 = x
  5. Write the solution. x is now on its own, so the equation tells us its value.
    x = 12

Check. Substitute x = 12 into the original equation. Left side: 123 + 2 = 4 + 2 = 6. Right side: 122 = 6. Both sides equal 6, so the solution is correct. ✓

Watch out. Multiply every term by the common denominator, including the terms that are not fractions. Then check that no fraction is left.

Watch out

Multiply every term. In x2 + 3 = 7, multiplying by 2 gives x + 6 = 14, not x + 3 = 14. And when the top has two terms, treat it like a bracket: 6 × x + 12 = 3(x + 1).

Higher: unknown in the denominator and two algebraic fractions. The same idea works when each fraction has an expression on the top, as in x + 12 − x − 24 = 4, and when two fractions are equal to each other. You may also meet equations where x is in the denominator, such as 12x = 3. Multiply both sides by the bottom to remove the fraction: 12 = 3x, so x = 4. Because we can never divide by zero, a solution is only allowed if it does not make a denominator equal to 0.

Worked example 9.6.5
Solve x + 12 − x − 24 = 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.

  1. Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
    x + 12 − x − 24 = 4
  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, 4
    Common denominator = 4
  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.
    4 × x + 12 = 2(x + 1)   (the bottom 2 goes into 4 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
    4 × x − 24 = (x − 2)   (the bottom 4 goes into 4 exactly 1 time, so the bottom disappears and the top is multiplied by 1)
    4 × 4 = 16   (a plain number or term with no bottom is simply multiplied by 4)
  4. Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
    2(x + 1) − (x − 2) = 16
  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(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.
    2x + 2 − x + 2 = 16
    x + 4 = 16
  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 + 4 = 16
    Now get the numbers on the other side. Subtract 4 from both sides: x = 12
  8. Read off the answer. x is already on its own.
    x = 12

Check. Put x = 12 back into the original equation. Left side: 13/2 − 10/4 = 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.

Worked example 9.6.6
Solve 3x + 14 = 2x + 23.

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.
    3x + 14 = 2x + 23
  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.
    4, 3
    Common denominator = 12
  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.
    12 × 3x + 14 = 3(3x + 1)   (the bottom 4 goes into 12 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
    12 × 2x + 23 = 4(2x + 2)   (the bottom 3 goes into 12 exactly 4 times, so the bottom disappears and the top is multiplied by 4)
  4. Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
    3(3x + 1) = 4(2x + 2)
  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 3(3x + 1):
    3 × 3x = 9x
    3 × 1 = 3
    so it is 9x + 3

    Expand 4(2x + 2):
    4 × 2x = 8x
    4 × 2 = 8
    so it is 8x + 8
  6. Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
    9x + 3 = 8x + 8
    9x + 3 = 8x + 8
  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: 9x + 3 = 8x + 8
    The x terms should be on one side. Subtract 8x from both sides: x + 3 = 8
    Now get the numbers on the other side. Subtract 3 from both sides: x = 5
  8. Read off the answer. x is already on its own.
    x = 5

Check. Put x = 5 back into the original equation. Left side: 16/4 = 4. Right side: 12/3 = 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.

Worked example 9.6.7
Solve 6x − 1 = 2.

Key words. The denominator is the bottom of a fraction. We cannot divide by zero, so a value of x that makes a denominator zero is never a solution.

Method. Multiply both sides by the bottom of the fraction to remove it, then solve the ordinary equation that is left.

  1. Write down the equation. The unknown is in the bottom (denominator) of a fraction. The bottom can never be zero, so we will check this at the end.
    6x − 1 = 2
  2. Multiply both sides by (x − 1). This removes the fraction. Multiplying by the bottom cancels it, because anything divided by itself is 1. We do it to both sides to keep the balance.
    6x − 1 × (x − 1) = 2 × (x − 1)
    6 = 2(x − 1)
  3. Expand the brackets. Multiply each term inside the bracket by the number outside it.
    Expand 2(x − 1):
    2 × x = 2x
    2 × (−1) = −2

    The equation is now 6 = 2x − 2
  4. Add 2 to both sides. This undoes the −2, because −2 + 2 = 0, which leaves the x term on its own. We do it to both sides to keep the balance.
    6 + 2 = 2x − 2 + 2
    8 = 2x
  5. Divide both sides by 2. 2x means 2 multiplied by x. The inverse of multiplying by 2 is dividing by 2, and we do it to both sides to keep the balance. This leaves x on its own.
    82 = 2x2
    x = 82 = 4
  6. Write the solution. x is now on its own, so the equation tells us its value.
    x = 4
  7. Check that the bottom is not zero. If a solution made the bottom equal to 0 the fraction would be impossible, so it would have to be rejected.
    x − 1 = 3 when x = 4, which is not 0, so the solution is allowed.

Check. Substitute x = 4: the bottom is 3, so the fraction is 63 = 2, which equals 2. ✓

Watch out. Multiply the whole of the other side by the bottom, so the right-hand side becomes a bracket. Never solve by "flipping" only part of it.

Exercise 9F  Equations with fractions

Non-calculator
1
FSolve these equations. Multiply both sides by the bottom number.
(a)x + 32 = 5
(b)x + 13 = 4
(c)x − 25 = 2
(d)x + 74 = 3
(e)x − 42 = 6
(f)x + 56 = 2
2
FSolve these equations. The top has two terms.
(a)2x + 13 = 5
(b)3x − 24 = 4
(c)5x + 32 = 14
(d)4x − 13 = 5
(e)2x + 57 = 3
(f)3x + 15 = 2
3
FSolve these equations. Find the lowest common multiple of the bottoms first.
(a)x2 + x3 = 10
(b)x3 + x4 = 7
(c)x2 − x5 = 6
(d)x4 + x2 = 9
(e)x3 + 2 = x2
(f)x5 + 3 = x2
4
FSolve these equations.
(a)x4 + 3 = 8
(b)x3 − 5 = 1
(c)x6 + 2 = 7
(d)x2 − 3 = 4
5
PSHalf of a number, plus 3, makes 11. Find the number.
6
PSAna has read a third of her book and 10 more pages. She has read 40 pages altogether. How many pages are in the book?
7
RKai solves x2 + 3 = 7. His first line is x + 3 = 14. Explain his mistake and find the correct solution.
8
PSThree numbers are x, x + 4 and x + 8. Their mean is 12. Find the three numbers.
9
RShow that x = 6 is the solution of x2 + x3 = 5.
10
FSolve these equations.
(a)x + 12 + x + 23 = 7
(b)x + 12 − x − 24 = 4
(c)3x − 22 − 3x − 33 = 3
(d)5x − 12 + 2x − 24 = 8
11
FSolve these equations. There is a fraction on each side.
(a)2x + 24 = 3x − 15
(b)3x + 14 = 2x + 23
(c)2x − 32 = 3x + 54
12
FSolve these equations. The unknown is in the bottom of the fraction.
(a)12x = 3
(b)6x − 1 = 2
(c)182x − 1 = 6
(d)15x + 2 = 3
13
RSolve 3x − 2 = 2x + 1.
14
RExplain why x = 2 cannot be a solution of 5x − 2 = 10, even though 5 ÷ (2 − 2) might look as though it works.

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