SAT Math skill page

SAT Rational Equations Practice

Clear denominators carefully, solve the resulting equation, and reject values that make a denominator zero.

Practice rational equations in Dolphin SAT Practice questions

12-18 min practice time3 questions on pageAdvanced Math

Practice time12-18 min

Question bank3 questions

Best forAdvanced Math

What this tests

What to know for this SAT skill

Identify values excluded by a denominator
Clear fractions using a common denominator
Solve equations containing rational expressions
Check that a solution is valid in the original equation

Dolphin question bank

Practice your skill

Question 1 of 3Hard

If f(

x) = \frac{2 x ^{2} - 8 x + 6}{x - 3}

, for which value of x is f(x) undefined?

Show answer and solution

Correct answerB. 3

How to solve it

Plan: Find excluded values from the original denominator before simplifying.

Set the denominator equal to zero: $x - 3 = 0$ , so $x = 3$ .
The numerator factors as $2 (x - 1) (x - 3)$ , but canceling $x - 3$ is valid only when $x \neq = 3$ .
Therefore, the original function is undefined at $x = 3$ , which is choice B.

CheckSubstituting $x = 3$ into the original denominator gives zero.

Common trapA canceled factor does not put an excluded value back into the original domain.

Question 2 of 3Hard

The equation

\frac{2}{x - 2} + \frac{3}{x + 5} = \frac{r x + t}{( x - 2 ) ( x + 5 )}

is true for all

x > 2

, where

r

and

t

are positive constants. What is the value of

r t

Show answer and solution

Correct answerC. 20

How to solve it

Plan: Combine the fractions over the stated common denominator, then match coefficients.

Rewrite the first numerator as $2 (x + 5)$ and the second as $3 (x - 2)$ .
Add: $2 (x + 5) + 3 (x - 2) = 2 x + 10 + 3 x - 6 = 5 x + 4$ .
So $r = 5$ and $t = 4$ . Therefore, $r t = 5 \cdot 4 = 20$ .

CheckFor $x > 2$ , neither denominator is zero, so the identity is valid on the stated domain.

Common trapDo not multiply the numerators together; adding fractions requires equivalent numerators over the common denominator.

Question 3 of 3Hard

What is the value of x that satisfies the equation

\frac{1}{x + 3} = \frac{2 x - 1}{x ^{2} - 9}

x = -3
x = -2
x = 2
x = 3

Show answer and solution

Correct answerB. x = -2

How to solve it

Plan: Record the excluded values, factor the denominator, and clear every denominator with the LCD.

The original denominators exclude $x = - 3$ and $x = 3$ .
Factor $x^{2} - 9 = (x - 3) (x + 3)$ and multiply both sides by $(x - 3) (x + 3)$ .
This gives $x - 3 = 2 x - 1$ , so $x = - 2$ .
Because $- 2$ is not excluded, it is a valid solution and corresponds to choice B.

CheckAt $x = - 2$ , both sides of the original equation equal $1$ .

Common trapDo not accept $x = \pm 3$ after cancellation; both make an original denominator zero.

Avoid these traps

Common mistakes on this skill

Forgetting excluded values

Any value that makes an original denominator zero cannot be a solution.

Clearing only one denominator

Multiply every term on both sides by the least common denominator.

Checking only the simplified equation

Substitute the result into the original rational equation to verify it remains defined.

Study plan

How to practice this skill in Dolphin

Write the values that make any denominator zero.
Find the least common denominator and multiply every term by it.
Solve the resulting polynomial or linear equation.
Reject excluded values and verify the remaining solution.

Practice rational equations in Dolphin SAT

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FAQ

Questions about SAT Rational Equations Practice

What is a rational equation?

It is an equation containing one or more fractions whose numerator or denominator includes an algebraic expression.

Why do rational equations have excluded values?

Division by zero is undefined, so values that make an original denominator zero are not allowed.

Should I cross-multiply every rational equation?

Cross-multiplication works for one fraction equal to one fraction. With more terms, clearing all denominators is usually safer.