Before clearing denominators, write down every value of x that makes a denominator zero. After solving, reject any answer that matches one of those values.
This is a core habit in algebraic equations and inequalities. The same thinking about allowed inputs appears in finding valid inputs for a rational expression, and in function domains generally.
Why does clearing denominators cause trouble?
Multiplying both sides by (x − 2) is only a legitimate step when x − 2 is not zero. If x = 2, you have multiplied by zero, and every equation becomes 0 = 0. So the new equation can have a root, x = 2, that the original never allowed.
If you write “x ≠ 2” before multiplying, you remember to test for it afterwards.
Method, step by step
- Factorise every denominator where you can.
- List the restrictions: set each denominator equal to zero and record those x values as excluded.
- Multiply through by the common denominator.
- Solve the resulting equation.
- Compare each solution to the restriction list and reject any that match.
- Substitute the survivors to confirm.
Worked example
Solve x²/(x − 2) = 4/(x − 2).
Step 1, restriction: x − 2 = 0 when x = 2, so x ≠ 2.
Step 2, multiply both sides by (x − 2): x² = 4.
Step 3, solve: x = 2 or x = −2.
Step 4, compare with the restriction: x = 2 is excluded. The value x = −2 is allowed.
Step 5, check x = −2: left side 4/(−4) = −1. Right side 4/(−4) = −1. Both agree.
Answer: x = −2 only.
The root x = 2 is not a slip in the algebra. It comes from multiplying by x − 2, which is zero at that value.
The mistake to watch for
The error is to give both solutions because the quadratic step looked correct.
Mistaken answer: x = 2 or x = −2
The student never wrote the restriction at the start, so nothing prompted a check at the end.
Another version is to cancel (x − 2) from both sides as if it were a number. That loses the solution set’s connection to the restriction and can hide a root. Keep the restriction written at the start of your working and read it once more before writing the final answer.
Check yourself
Try these, then open each answer.
1. Solve (x + 1)/(x² − 9) = 2/(x − 3).
Show answer
x² − 9 = (x − 3)(x + 3), so restrictions are x ≠ 3 and x ≠ −3. Multiply by (x − 3)(x + 3): x + 1 = 2(x + 3) = 2x + 6, so x = −5.
x = −5 is allowed. Check: (−4)/(25 − 9) = −4/16 = −1/4, and 2/(−8) = −1/4.
x = −5
2. Solve 2x/(x − 3) = 6/(x − 3) + 1.
Show answer
Restriction: x ≠ 3. Multiply by (x − 3): 2x = 6 + (x − 3), so x = 3. That is the restricted value, so it is rejected.
No solution
3. Solve 1/x + 1/(x + 1) = 5/6.
Show answer
Restrictions: x ≠ 0 and x ≠ −1. Multiply by 6x(x + 1): 6(x + 1) + 6x = 5x(x + 1), so 12x + 6 = 5x² + 5x, giving 5x² − 7x − 6 = 0, which factorises as (5x + 3)(x − 2) = 0.
Roots: x = 2 and x = −3/5. Neither is restricted. Check x = 2: 1/2 + 1/3 = 5/6. Check x = −3/5: −5/3 + 5/2 = 5/6.
x = 2 or x = −3/5
Where this leads next
The idea of unwanted roots after an algebraic step links back to rejecting an extraneous root after squaring. When ready, try the mixed practice set, and use the non-calculator working trainer for the fraction checks.
Setting out restrictions is a small habit that protects a lot of marks. If you would like to build it into your own working with feedback as you go, our teachers can help in online one-to-one Additional Mathematics tuition.