Year 11 Extended · Unit 4 Rational expressions and functions · Lesson 6 of 7

What happens when you check the answer to a fraction equation?

An equation with letters in the bottoms can be cleared of fractions by multiplying through. This lesson clears them, solves the linear or quadratic equation that results, and checks each answer in the original.

FunctionsNo calculator

Warm-up

Three quick ones from earlier work. Try each in your head or on paper before you open the answer.

  1. From Year 9: solve x2 − x − 6 = 0.
    Answerx = 3 or x = −2.
  2. From Year 8: solve x⁄4 = 3⁄10.
    Answerx = 1.2.
  3. From lesson 3: write 1⁄x + 1⁄x + 1 as one fraction.
    Answer(2x + 1) ÷ (x(x + 1)).
Words used in this lesson
clear the fractions
multiply every term so that no bottoms are left. For example: multiply by x(x + 2).
reject
decide a value is not allowed. For example: a negative length is rejected.
root
a value of x that solves an equation. For example: the roots of x2 = 9 are 3 and −3.
extraneous solution
a value that the algebra gives but the original does not allow. For example: a value that makes a bottom 0.
rate
how much is done in one unit of time. For example: 30 pages printed each minute.

When does a correct step give a wrong answer?

Naledi solves (x2 − 4) ÷ (x − 2) = 4. She cancels x − 2, writes x + 2 = 4, and finds x = 2. Then she checks her answer by putting x = 2 back into the original equation.

Predict first: what do you think she finds when she substitutes? Then use the figure to move the horizontal line and see what it meets.

−1
x = −3the solution
x + 2 = −1after cancelling

(x² − 4)/(x − 2) = −1: after cancelling, x + 2 = −1, so x = −3, which is allowed.

Show me the steps
  1. c = −1: (x2 − 4)/(x − 2) = −1 cancels to x + 2 = −1, so x = −3. The bottom is −5, not 0, so x = −3 is a solution.
  2. c = 4: x + 2 = 4 gives x = 2. But x = 2 makes the bottom 0, so it is rejected: the horizontal line passes through the gap and there is no solution.
  3. c = 6: x + 2 = 6 gives x = 4. The bottom is 2, so x = 4 is a solution, and the line meets the graph at (4, 6).

Clear the fractions, solve, then check. Multiply every term by the common denominator so that no bottoms are left. The equation that results may be linear or quadratic. Solve it fully.

Then reject any value that makes a denominator of the original equation zero. What is left, even if it is nothing, is the solution set.

Check in the original, not in the last line. A value can satisfy the cleared equation and still not be a solution. Substitute into the original: the bottom is x − 2, which is 0 at x = 2, so the left side is 0/0, which has no value.

Worked examples

1. Solve 3/(x + 1) = 2/(x − 2).

  1. Multiply both sides by (x + 1)(x − 2): 3(x − 2) = 2(x + 1).
  2. 3x − 6 = 2x + 2, so x = 8. Neither bottom is 0 at 8.

2. Solve 2/x + 3/(x + 2) = 1.

  1. Multiply by x(x + 2): 2(x + 2) + 3x = x(x + 2), so 5x + 4 = x2 + 2x.
  2. x2 − 3x − 4 = 0 factorizes as (x − 4)(x + 1) = 0, so x = 4 or x = −1. Neither is 0 or −2.

3. Solve (x2 + x)/(x − 3) = 12/(x − 3).

  1. Multiply by x − 3: x2 + x = 12, so x2 + x − 12 = (x + 4)(x − 3) = 0.
  2. x = 3 makes the bottom 0, so it is rejected. The only solution is x = −4.

Your go. The first step is given. Solve 4/(x + 2) = 3/(x − 1). Multiply both sides by (x + 2)(x − 1).

  1. Write the equation without fractions.
  2. Solve it and check against the excluded values.
Check the answer

4(x − 1) = 3(x + 2), so 4x − 4 = 3x + 6 and x = 10. The bottoms are 12 and 9, not 0.

Check before you go on

1. Solve x ÷ (x − 3) = 3 ÷ (x − 3) + 2. Which statement is right?

Your turn

Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.

Level 1 · Getting started

2. Solve 5/(x + 2) = 2/(x − 1).

3. Solve 12/x − 3 = 1.

Level 2 · Building

4. The equation 3/x + 4/(x + 2) = 1 has two solutions. Find the positive one.

5. Solve x2/(x − 4) = 16/(x − 4). Give the only solution.

Level 3 · Stretch

6. 1/(x − 1) + 1/(x + 1) = 3/4. Find the positive solution.

7. Tap A fills a tank alone in x hours and tap B alone in x + 4 hours. Together they fill 3/8 of the tank each hour. Find x.

Something to think about

Are there fraction equations with a value rejected and a second solution that stays? Invent one, then check it by putting both values back into the original equation.

One way to think about it

Yes. For example (x2 + 2x)/(x − 5) = 35/(x − 5) gives x2 + 2x − 35 = (x + 7)(x − 5) = 0, where x = 5 is rejected and x = −7 stays. Always substitute the surviving value into the original equation.

Exit ticket

8. Which is the solution of x ÷ (x − 3) + 2 ÷ (x + 3) = 18 ÷ (x2 − 9)?