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College Algebra Quiz

College Algebra Quiz: Solve Rational Equations

Practice Solve Rational Equations in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 1

0 of 1 answered

A student attempts to solve x+2x−3=x−1x+1+8(x−3)(x+1)\frac{x+2}{x-3} = \frac{x-1}{x+1} + \frac{8}{(x-3)(x+1)}x−3x+2​=x+1x−1​+(x−3)(x+1)8​ and finds that x=3x = 3x=3 and x=−1x = -1x=−1. What should the student conclude?

Select an answer to continue

What this quiz covers

This quiz focuses on Solve Rational Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A student attempts to solve x+2x−3=x−1x+1+8(x−3)(x+1)\frac{x+2}{x-3} = \frac{x-1}{x+1} + \frac{8}{(x-3)(x+1)}x−3x+2​=x+1x−1​+(x−3)(x+1)8​ and finds that x=3x = 3x=3 and x=−1x = -1x=−1. What should the student conclude?

  1. Both solutions are valid since they satisfy the algebraic equation
  2. Only x=3x = 3x=3 is valid; x=−1x = -1x=−1 is extraneous due to domain restrictions
  3. Only x=−1x = -1x=−1 is valid; x=3x = 3x=3 is extraneous due to domain restrictions
  4. Both solutions are extraneous due to domain restrictions, so there is no solution (correct answer)

Explanation: The original equation has denominators x−3x-3x−3, x+1x+1x+1, and (x−3)(x+1)(x-3)(x+1)(x−3)(x+1). This means the domain excludes x=3x = 3x=3 and x=−1x = -1x=−1. Since both algebraic solutions x=3x = 3x=3 and x=−1x = -1x=−1 make denominators zero in the original equation, both are extraneous solutions that must be rejected. Therefore, the equation has no solution. This illustrates the critical importance of checking solutions against domain restrictions when solving rational equations.