How do I check a solution to a rational equation?

The quick answer

Before you cancel a factor or clear fractions, write down every value that makes an original denominator zero. Solve the equation, cross out any excluded candidates, then substitute the remaining candidates into the original equation. Keep each value that gives two defined, equal sides.

Would you accept 7 as the answer?

Imagine finding this line on your scratch paper during entrance exam practice:

$x + 7 = 14$

You circle $x = 7$. The arithmetic works. Now look at the question you started with:

$\frac{x^2 - 49}{x - 7} = 14$

Put 7 into the original fraction. The top becomes 0, and the bottom becomes 0. The expression $0/0$ is undefined. Your last line admits an input that the first line excludes. Where did that happen?

Find the condition hidden in the cancellation

Factor the numerator as $(x - 7)(x + 7)$. The denominator is $x - 7$, so write $x \ne 7$ beside the equation. For every allowed input, the common factor cancels and the fraction equals $x + 7$. At 7, canceling would require dividing by zero. OpenStax explains this restriction when simplifying fractions with variables. Read the section on rational expressions.

Return to $x + 7 = 14$. Its only candidate is 7, and 7 is excluded. The original equation therefore has no solution. That is a complete answer. There is no need to find another number to fill the blank.

Change one number and try again

Keep the fraction and change the right side from 14 to 12. Predict whether the restriction changes.

$\frac{x^2 - 49}{x - 7} = 12$

The original denominator still rules out 7. After cancellation, $x + 7 = 12$ gives $x = 5$. Check 5 in the original fraction, keeping the signs on both numbers:

$\frac{25 - 49}{5 - 7} = \frac{-24}{-2} = 12.$

The solution is 5. Both questions use the same simplification. The difference is where each candidate lands. In the first problem the candidate made the denominator zero. In the second it gave a defined fraction equal to the requested value.

A scratch-paper checklist

When you see a variable below a fraction bar, leave room beside the problem for three short lines. This also works when you multiply both sides by a common denominator. OpenStax walks through solving rational equations.

1. Excluded values. Set each original denominator equal to zero and list the values you find. For a denominator like $(x - 2)(x + 5)$, that line reads “exclude 2 and −5.”
2. Candidates. Solve using allowed operations. Cancel whole factors. If a numerator contains a sum, factor it first to see whether it actually shares a factor with the denominator.
3. Original check. Cross out excluded candidates. For each candidate left, substitute into the first equation and evaluate both sides. A negative denominator is fine. A zero denominator ends the check for that candidate.

Three questions to try on paper

Solve over the real numbers. Write the excluded values first, then the solution set. Cover the answers until you have both lines for each problem.

1. $\dfrac{(x-3)(x+4)}{x-3}=0$
2. $\dfrac{x^2-16}{x+4}=-8$
3. $\dfrac{(x-2)(x+1)}{x-2}=x+1$

Check your reasoning

1. Exclude 3. The solution is −4. The simplified equation is $x+4=0$. At −4 the original numerator is $(-7)(0)=0$ and the denominator is −7. Zero divided by −7 equals zero. If you picked 3 too, check what happens below the fraction bar at 3.

2. Exclude −4. There is no solution. Factor the numerator as $(x-4)(x+4)$. The simplified equation $x-4=-8$ gives the single candidate −4. That candidate produces $0/0$ in the original problem.

3. Every real number except 2 is a solution. For any allowed input, the factor $x-2$ cancels and both sides equal $x+1$. You can spot-check with $x=0$, which gives 1 on each side. The algebra explains why the equality holds for every allowed input. At 2, the left side is undefined.

If you missed one, put the two relevant lines side by side in your mistake log. For question 2, write “candidate −4” beside “original denominator −4 + 4 = 0.” On your next review day, hide those lines and reconstruct them using the active recall routine. You can use the same three-line check when reviewing algebra in the free PasadoPH diagnostic.

Common questions

Can zero be a valid solution to a rational equation?

Yes. For example, x divided by (x + 2) equals zero at x = 0. The numerator is zero and the denominator is 2, so the fraction is defined and equals zero.

Why do I keep the restriction after canceling a factor?

Canceling divides the numerator and denominator by the same factor. That division requires the factor to be nonzero. The simplified expression agrees with the original only for inputs the original expression allows.

What if both sides simplify to the same expression?

If every step is valid for the allowed inputs, the equation holds for every one of those inputs. Report the solution set together with the original restrictions.

Sources & further reading

Sources checked for this guide on November 6, 2026. Admissions requirements can change; confirm your application cycle with the university.

• OpenStax — Intermediate Algebra 2e, Multiply and Divide Rational Expressions
• OpenStax — Intermediate Algebra 2e, Solve Rational Equations