Why must I check that the denominator is nonzero?

A rational expression is undefined when its denominator is zero. If your solution makes Q(x) = 0, then the original equation does not exist at that point, so it cannot be a valid solution.

Can I always use the method of multiplying both sides by the denominator?

Yes, multiplying both sides of \frac{P(x)}{Q(x)} = 0 by Q(x) gives P(x) = 0, which is a valid step. However, you must still check that your solutions do not make Q(x) = 0 in the original equation.

What if the numerator is a quadratic that does not factor nicely?

You can use the quadratic formula or other methods to solve P(x) = 0. Even if the solutions are irrational, you still must check whether they make the denominator zero.

What do I do if the numerator and denominator have a common factor?

You can cancel the common factor to simplify, but remember that the original domain restriction still applies. For example, \frac{x(x+1)}{x} = 0 simplifies to x + 1 = 0 (for x \neq 0), giving x = -1.

Rational Equations

A rational equation has a polynomial fraction equal to zero; solve by setting the numerator to zero, then verify the denominator is nonzero.

\frac{P(x)}{Q(x)} = 0 \implies P(x) = 0, ; Q(x) \neq 0

Solve a problem with Rational Equations

Type the problem. The solver will use Rational Equations where Rational Equations is the right tool, and tell you when it is not.

What each symbol means

What Rational Equations takes

Symbol	Meaning
$P$	The numerator polynomial in the equation $\frac{P(x)}{Q(x)} = 0$ ; if this is misread as the whole equation, you lose the crucial denominator restriction.
$Q$	The denominator polynomial in the equation $\frac{P(x)}{Q(x)} = 0$ ; it must never equal zero, or the entire fraction becomes undefined.
$x$	The unknown variable you are solving for; solutions must be verified to ensure they do not make the denominator zero.

When to use it

When you have a polynomial fraction set equal to zero and need to find the values of the variable that make it zero.

Level

Usually taught in: Algebra II

Worked examples

1. Solve a simple rational equation

Problem

Solve

\frac{x - 3}{x + 2} = 0

for

x

$\frac{x - 3}{x + 2} = 0$
A fraction equals zero only when its numerator is zero and the denominator is nonzero.
$x - 3 = 0$
Set the numerator equal to zero, since that's the only way the fraction can be zero.
$x = 3$
Add 3 to both sides to isolate $x$ .
$3 + 2 = 5 \neq 0$
Substitute $x = 3$ into the denominator to verify it is nonzero.

Answer: $x = 3$

A fraction equals zero only when its numerator is zero and the denominator is nonzero. We set the numerator to zero and solve, then verify our answer doesn't make the denominator zero.

2. Solve a rational equation with multiple candidates

Problem

Solve

\frac{x(x - 2)}{x - 2} = 0

for

x

$\frac{x(x - 2)}{x - 2} = 0$
Start with the equation. A fraction equals zero only when its numerator is zero and its denominator is nonzero.
$x(x - 2) = 0$
Set the numerator equal to zero.
$x = 0 \quad \text{or} \quad x = 2$
Apply the zero product property: either $x = 0$ or $x - 2 = 0$ .
$x = 0: \quad 0 - 2 = -2 \neq 0$
Substitute $x = 0$ into the denominator to check: $0 - 2 = -2$ is nonzero, so $x = 0$ is valid.
$x = 2: \quad 2 - 2 = 0$
Substitute $x = 2$ into the denominator: $2 - 2 = 0$ , making the fraction undefined, so $x = 2$ is not valid.

Answer: $x = 0$

When the numerator factors, we get multiple candidate solutions from the zero product property. However, we must reject any solution that makes the denominator zero, even if it appears in the factored numerator. Only $x = 0$ is valid here.

3. Solve a rational equation from a word problem about concert tickets

Problem

A concert venue offers general admission tickets at $60 and VIP tickets at $120. The net profit model is

\frac{2x - 150}{x + 30}

where

x

is the number of general admission tickets sold. If the net profit is zero, how many general admission tickets were sold?

$\frac{2x - 150}{x + 30} = 0$
Set up the equation from the problem statement.
$2x - 150 = 0$
Set the numerator equal to zero, since a fraction equals zero only when its numerator is zero and its denominator is nonzero.
$2x = 150$
Add 150 to both sides.
$x = 75$
Divide both sides by 2.
$75 + 30 = 105 \neq 0$
Verify the denominator is nonzero by substituting $x = 75$ into $x + 30$ .

Answer: $x = 75 \text{ general admission tickets}$

In real-world applications like concert revenue, rational expressions often describe break-even points where profit is zero. We set the numerator equal to zero, solve for the variable, and verify the denominator is nonzero to ensure our answer is valid.

Common mistakes

Where Rational Equations usually goes wrong

The mistake	Why it is wrong	The fix
After solving $\frac{x(x - 2)}{x} = 0$ , student writes $x = 0$ or $x = 2$ as the final answer without checking the denominator.	When $x = 0$ , the denominator becomes zero, making the fraction undefined, so $x = 0$ is not a valid solution.	Always verify that solutions do not make the denominator zero; in this case, only $x = 2$ is valid.
Student sees $\frac{(x+1)(x-3)}{x-3} = 0$ and concludes $x = -1$ or $x = 3$ as the solution.	The zero product property applies only to the numerator. The denominator must never be zero, so $x = 3$ violates the domain restriction even though it appears to solve the equation.	Set only the numerator to zero: $(x+1)(x-3) = 0$ gives $x = -1$ or $x = 3$ ; then reject $x = 3$ because it makes the denominator zero.
Student solves $\frac{x^2 - 1}{x - 1} = 0$ by multiplying both sides by $(x - 1)$ to get $x^2 - 1 = 0$ , then factors to $(x-1)(x+1) = 0$ and writes $x = 1$ or $x = -1$ as the answer.	When clearing fractions, the domain restriction $x \neq 1$ persists throughout. The solution $x = 1$ violates this restriction, even though it satisfies the simplified equation.	After clearing fractions, check that any solutions obtained do not violate the original domain restrictions; here, only $x = -1$ is valid.

Tips and when to use something else

Always verify that your solutions do not make the denominator zero; this is the most common error in rational equations.
To find potential solutions, set the numerator $P(x) = 0$ and solve; you can use Solving Quadratics by Factoring if the numerator is a quadratic.
Check for extraneous solutions (solutions that make the denominator zero) after solving; learn more in Extraneous Solutions.
If the fraction can be simplified by canceling common factors, do so first, but remember that the domain restriction still applies.

Related formulas

Linear EquationSolve for the value of x in any linear equation by rearranging to isolate x, finding the unique solution where the expression equals zero.Zero Product PropertyA product equals zero if and only if at least one factor equals zero; use this property to solve equations by factoring.Extraneous SolutionsA solution to the transformed equation that doesn't satisfy the original; found when squaring both sides or applying other non-reversible operations.Solving Quadratics by FactoringSolve quadratic equations by factoring the polynomial into binomials, then apply the Zero Product Property to find the solutions.Simplifying FractionsSimplify a fraction to its lowest terms by dividing both the numerator and the denominator by their greatest common factor.

Frequently asked questions

Why must I check that the denominator is nonzero?: A rational expression is undefined when its denominator is zero. If your solution makes $Q(x) = 0$ , then the original equation does not exist at that point, so it cannot be a valid solution.
Can I always use the method of multiplying both sides by the denominator?: Yes, multiplying both sides of $\frac{P(x)}{Q(x)} = 0$ by $Q(x)$ gives $P(x) = 0$ , which is a valid step. However, you must still check that your solutions do not make $Q(x) = 0$ in the original equation.
What if the numerator is a quadratic that does not factor nicely?: You can use the quadratic formula or other methods to solve $P(x) = 0$ . Even if the solutions are irrational, you still must check whether they make the denominator zero.
What do I do if the numerator and denominator have a common factor?: You can cancel the common factor to simplify, but remember that the original domain restriction still applies. For example, $\frac{x(x+1)}{x} = 0$ simplifies to $x + 1 = 0$ (for $x \neq 0$ ), giving $x = -1$ .

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Reviewed 2026-09-18