How do you solve a rational equation?

List the values that make a denominator 0, multiply every term by the least common denominator to clear the fractions, solve, and reject any answer on the restriction list.

What is an extraneous solution?

A value that solves the equation after the fractions are cleared but makes a denominator 0 in the original equation, so it is not a real solution.

When does a rational equation have no solution?

When every value you find is extraneous, or when the variable cancels and leaves a false statement.

How many solutions does 2x/(x - 4) = 8/(x - 4) + 3 have?

Zero. The only candidate, x = 4, makes the denominators 0.

What are all the solutions to x/(x - 1) - 2/(x + 1) = 4/(x^2 - 1)?

x = 2 only. x = -1 is extraneous.

Solving Rational Equations

SAT Math · Advanced Math · Updated October 2026

A rational equation has the variable in a denominator. You may be asked to solve one, to say how many solutions it has, to find a missing constant from a given solution, or to find the value that leaves no solution at all.

Before you solve anything

Write down the values of $x$ that make a denominator $0$. Those values can never be answers, no matter what your algebra says later.

Then clear the fractions by multiplying every term by the least common denominator, solve what is left, and cross off any answer that is on your list. That crossed-off answer is called an extraneous solution, and the SAT usually includes it among the answer choices.

Worked examples

One fraction on each side

Solve $\dfrac{3}{x - 2} = \dfrac{6}{x + 1}$.

Restrictions: $x \neq 2$ and $x \neq -1$.
Cross-multiply: $3(x + 1) = 6(x - 2)$, so $3x + 3 = 6x - 12$.
$15 = 3x$, so $x = 5$. It is not on the restriction list, so $x = 5$ is the solution. Check: $\dfrac{3}{3} = 1$ and $\dfrac{6}{6} = 1$.

The answer that is not allowed

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

Restriction: $x \neq 3$.
Multiply every term by $x - 3$: $x = 3 + 2(x - 3)$, so $x = 2x - 3$.
That gives $x = 3$, which is on the restriction list. The equation has no solution.

A quadratic with one extraneous root

Solve $\dfrac{x}{x - 1} - \dfrac{2}{x + 1} = \dfrac{4}{x^2 - 1}$.

Factor $x^2 - 1 = (x - 1)(x + 1)$. Restrictions: $x \neq 1$ and $x \neq -1$.
Multiply every term by $(x - 1)(x + 1)$: $x(x + 1) - 2(x - 1) = 4$.
Expand: $x^2 + x - 2x + 2 = 4$, so $x^2 - x - 2 = 0$ and $(x - 2)(x + 1) = 0$.
$x = 2$ or $x = -1$. Cross off $x = -1$. The only solution is $x = 2$.

Finding a constant from a given solution

The equation $\dfrac{k}{x + 2} = \dfrac{3}{x - 1}$ has the solution $x = 4$. What is $k$?

Substitute $x = 4$: $\dfrac{k}{6} = \dfrac{3}{3}$.
The right side is $1$, so $\dfrac{k}{6} = 1$ and $k = 6$.
Substituting first is faster than solving for $x$ in terms of $k$.

How these questions are asked

Step	What to do
1. List the restrictions	Set each denominator equal to $0$. Factor first: $x^2 - 9 = (x - 3)(x + 3)$ rules out $x = 3$ and $x = -3$.
2. Clear the fractions	With one fraction on each side, cross-multiply. Otherwise multiply every term on both sides by the least common denominator.
3. Solve	What is left is a linear or quadratic equation. Solve it the usual way.
4. Check	Cross off any answer from your restriction list. If every answer is crossed off, the equation has no solution.

Mistakes I see often

Skipping the check. An answer that makes any denominator $0$ is not a solution, even though your algebra produced it.

Multiplying only some terms by the denominator. Every term on both sides gets multiplied, including whole numbers like the $2$ in Example 2.

Dropping the minus sign across a numerator. $-\dfrac{x - 3}{x + 1}$ becomes $-(x - 3) = -x + 3$ after clearing, not $-x - 3$.

Cancelling across addition. In $\dfrac{x + 4}{x}$ you cannot cancel the $x$ terms. You can only cancel factors.

Not factoring the denominator. $x^2 - 4$ is $(x - 2)(x + 2)$. If you miss this, you miss a restriction and the least common denominator.

Try these

Work each question on paper, and check every answer against the denominators before you open the solution.

Easy

What value of $x$ satisfies $\dfrac{4}{x} = \dfrac{2}{5}$?

A$\dfrac{8}{5}$
B$10$
C$\dfrac{5}{2}$
D$20$

Show solution

Answer: B. Cross-multiply: $2x = 20$, so $x = 10$. Check: $\dfrac{4}{10} = \dfrac{2}{5}$.

Easy

What is the solution to $\dfrac{2}{x + 3} = \dfrac{1}{x - 1}$?

A$x = 1$
B$x = -3$
C$x = 5$
D$x = -5$

Show solution

Answer: C. Cross-multiply: $2(x - 1) = x + 3$, so $2x - 2 = x + 3$ and $x = 5$. It is not $-3$ or $1$, so it is allowed. Check: $\dfrac{2}{8} = \dfrac{1}{4}$.

MediumStudent-produced response

What is the solution to $\dfrac{1}{x} + \dfrac{1}{2x} = \dfrac{3}{4}$?

Enter your answer.

Show solution

Answer: $2$. Combine the left side over $2x$: $\dfrac{2}{2x} + \dfrac{1}{2x} = \dfrac{3}{2x}$. Then $\dfrac{3}{2x} = \dfrac{3}{4}$, so $2x = 4$ and $x = 2$.

Medium

How many solutions does the equation $\dfrac{2x}{x - 4} = \dfrac{8}{x - 4} + 3$ have?

AZero
BExactly one
CExactly two
DInfinitely many

Show solution

Answer: A. The restriction is $x \neq 4$. Multiply every term by $x - 4$: $2x = 8 + 3(x - 4) = 3x - 4$, so $x = 4$. That value makes the denominators $0$, so it is extraneous and the equation has no solution.

MediumStudent-produced response

The equation $\dfrac{k}{x + 2} = \dfrac{3}{x - 1}$ has the solution $x = 4$, where $k$ is a constant. What is the value of $k$?

Enter your answer.

Show solution

Answer: $6$. Substitute $x = 4$: $\dfrac{k}{6} = \dfrac{3}{3} = 1$, so $k = 6$.

MediumStudent-produced response

The equation $\dfrac{k}{x^2 - 4} = \dfrac{3}{x - 2}$ has the solution $x = 5$, where $k$ is a constant. What is the value of $k$?

Enter your answer.

Show solution

Answer: $21$. Substitute $x = 5$: $\dfrac{k}{25 - 4} = \dfrac{3}{3}$, so $\dfrac{k}{21} = 1$ and $k = 21$.

Hard

What are all the solutions to $\dfrac{x}{x - 1} - \dfrac{2}{x + 1} = \dfrac{4}{x^2 - 1}$?

A$x = -1$ only
B$x = 2$ and $x = -1$
CNo solution
D$x = 2$ only

Show solution

Answer: D. Multiply every term by $(x - 1)(x + 1)$: $x(x + 1) - 2(x - 1) = 4$, which simplifies to $x^2 - x - 2 = 0$, or $(x - 2)(x + 1) = 0$. The value $x = -1$ makes two denominators $0$, so it is extraneous. Only $x = 2$ works.

Hard

For what value of the constant $a$ does the equation $\dfrac{x + a}{x - 2} = 3$ have no solution?

A$-6$
B$-2$
C$2$
D$6$

Show solution

Answer: B. Multiply by $x - 2$: $x + a = 3x - 6$, so $x = \dfrac{a + 6}{2}$. That is a valid solution unless it equals the restricted value $2$. Setting $\dfrac{a + 6}{2} = 2$ gives $a = -2$. Check: with $a = -2$, the left side is $\dfrac{x - 2}{x - 2} = 1$, which never equals $3$.

HardStudent-produced response

The equation $\dfrac{2}{x - k} + \dfrac{3}{x + k} = \dfrac{k + 11}{x^2 - k^2}$ has the solution $x = 4$, where $k$ is a positive constant. What is the value of $k$?

Enter your answer.

Show solution

Answer: $4.5$ (or $\dfrac{9}{2}$). Multiply every term by $x^2 - k^2 = (x - k)(x + k)$: $2(x + k) + 3(x - k) = k + 11$, so $5x - k = k + 11$ and $5x = 2k + 11$. With $x = 4$: $20 = 2k + 11$, so $k = 4.5$. Check that $x = 4$ is allowed: it is not $\pm 4.5$.

Hard

Pipe A can fill a tank in $6$ hours. Pipe B can fill the same tank in $3$ hours. Working together, the pipes fill the tank in $t$ hours, where $\dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{t}$. What is the value of $t$?

A$4.5$
B$9$
C$2$
D$1.5$

Show solution

Answer: C. $\dfrac{1}{6} + \dfrac{2}{6} = \dfrac{3}{6} = \dfrac{1}{2}$, so $\dfrac{1}{t} = \dfrac{1}{2}$ and $t = 2$. Choice A is the average of $6$ and $3$, but working together must be faster than the faster pipe alone.

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