Equations with algebraic fractions and rational coefficients | Year 10

01 / LEARN

The idea, step by step

Record, clear, solve

First identify denominator restrictions. Multiply every term on both sides by a common denominator, then solve the resulting equation. Do not multiply just the fractional terms. Keep the resulting value as a candidate until it passes the restrictions and the original equation.

Why a candidate may fail

Multiplying by an expression can hide an undefined original value. If a candidate makes an original denominator zero, it is not a solution. A contradiction means no solution; an identity means all values allowed by the original restrictions. These conclusions are part of solving, not optional final decoration.

A common trap

Accepting $x = 2$ after cancelling a denominator $x - 2$ .

Instead: The original expression is undefined at $2$ . Restrictions must be checked even when the simplified equation has that root.

02 / WORKED EXAMPLES

See the thinking

Example 1

Two algebraic denominators

Solve $\frac{3}{x - 1} = \frac{2}{x + 1}$ .

Restrictions: $x \neq 1, - 1$ .
Multiply by $(x - 1) (x + 1)$ : $3 (x + 1) = 2 (x - 1)$ .
Expand: $3 x + 3 = 2 x - 2$ , so $x = - 5$ .
Check: both original sides equal $- 1 / 2$ .

x = - 5

Example 2

An excluded candidate

Solve $\frac{x + 1}{x - 2} = \frac{3}{x - 2}$ .

The original fractions require $x \neq 2$ .
Multiplying by $x - 2$ gives $x + 1 = 3$ , so the candidate is $x = 2$ .
That candidate is excluded and makes both original fractions undefined.

No solution

Check your understanding

Solve $\frac{2}{x + 1} = 1$ .

Show answer

$x = 1$

The restriction is $x \neq - 1$ . Multiplication gives $2 = x + 1$ , and $x = 1$ passes the original equation.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $\frac{x}{3} + \frac{1}{2} = \frac{7}{6}$ .
2 marks
Solve $\frac{5}{x - 2} = 1$ .
2 marks
Solve $\frac{2}{x + 1} = \frac{3}{x + 4}$ .
2 marks
Solve $\frac{x - 1}{x + 2} = \frac{- 3}{x + 2}$ .
2 marks
A cost of 72 dollars is shared equally. Adding two people reduces each share from $72 / n$ to $72 / (n + 2)$ . If the original share is 12 dollars, find $n$ and the new share.
3 marks
Solve $\frac{2 x + 4}{x + 2} = 2$ and state the full solution set.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Solve $\frac{x}{3} + \frac{1}{2} = \frac{7}{6}$ .
2 marks
Multiply every term by $6$ : $2 x + 3 = 7$ .
$x = 2$
Solve $\frac{5}{x - 2} = 1$ .
2 marks
$x \neq 2$ ; multiply by $x - 2$ to get $5 = x - 2$ .
Check $5 / (7 - 2) = 1$ .
$x = 7$
Solve $\frac{2}{x + 1} = \frac{3}{x + 4}$ .
2 marks
$x \neq - 1, - 4$ ; $2 (x + 4) = 3 (x + 1)$ .
This gives $2 x + 8 = 3 x + 3$ , hence $x = 5$ .
$x = 5$
Solve $\frac{x - 1}{x + 2} = \frac{- 3}{x + 2}$ .
2 marks
$x \neq - 2$ . Clearing gives $x - 1 = - 3$ , hence $x = - 2$ .
The candidate is excluded.
No solution
A cost of 72 dollars is shared equally. Adding two people reduces each share from $72 / n$ to $72 / (n + 2)$ . If the original share is 12 dollars, find $n$ and the new share.
3 marks
$72 / n = 12$ gives $n = 6$ , a positive whole number.
The new group has $8$ people, so the new share is $72 / 8 = 9$ .
$n = 6$ ; new share $9$ .
Solve $\frac{2 x + 4}{x + 2} = 2$ and state the full solution set.
3 marks
The denominator requires $x \neq - 2$ .
The numerator is $2 (x + 2)$ , so both sides are $2$ for every allowed value.
All real $x$ except $- 2$ .

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Curriculum reference & lesson scope

Australian Curriculum v9 and South Australian Mathematics Curriculum

Year 10 algebra, including the approved fuller school-course treatment of rational expressions and formula restrictions; quadratic methods are in their own chapter.

Official curriculum source ↗

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