Solving rational equations

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $\dfrac{6}{x} = 2$ . Which value works?

$x = 3$
$x = 12$

Key idea

Clear the denominators, solve, then check.

Multiply both sides by the common denominator, solve the resulting polynomial equation, then check every solution in the original equation: a value that makes a denominator 0 is extraneous and must be rejected.

Worked example

$\dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 1$ .

x = 2 + (x - 2)

, so

x = x

Multiply by x − 2.

…except where the denominator is 0

True for every x.

Solutions: every

x \ne 2

x = 2 is excluded.

Your turn, guided

Your turn: $\dfrac{x}{x + 1} = \dfrac{2}{3}$ .

x = 2

3x = 2x + 2.

Practice ★

Solve $\dfrac{x}{x + 2} = \dfrac{ 2 }{ 7 }$ .
Solve $\dfrac{x}{x + 2} = \dfrac{ 4 }{ 6 }$ .
Solve $\dfrac{x}{x + 6} = \dfrac{ 5 }{ 8 }$ .

Watch out!

Practice ★★

Solve $\dfrac{1}{x} + \dfrac{1}{ 11 } = \dfrac{1}{ 5 }$ .
Solve $\dfrac{1}{x} + \dfrac{1}{ 12 } = \dfrac{1}{ 2 }$ .

Solve $\dfrac{x^2 - 4}{x - 2} = 4$ .
1. No solution: simplifying gives x = 2, but x = 2 makes the denominator 0
2. x = 2
3. Every x except 2

Exit ticket

Three questions to check you have got it.

Solve $\dfrac{x}{x + 2} = \dfrac{ 2 }{ 7 }$ .
Solve $\dfrac{x}{x + 2} = \dfrac{ 4 }{ 6 }$ .
Solve $\dfrac{x}{x + 6} = \dfrac{ 5 }{ 8 }$ .

Summary card

Rational equations

Clear denominators → solve → check in the original.

Watch out: reject values that make a denominator 0.

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