Math 3 Quiz: Solving Rational Equations
6 questions · exam conditions
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Solving Rational EquationsQuestion 1 of 6

To solve 3x−1−2x+2=x+7(x−1)(x+2)\frac{3}{x-1} - \frac{2}{x+2} = \frac{x+7}{(x-1)(x+2)}, a student multiplies both sides by (x−1)(x+2)(x-1)(x+2) and gets 3(x+2)−2(x−1)=x+73(x+2) - 2(x-1) = x+7. What is the next step, and what should the student be careful about?

Expand to get 3x+6−2x+2=x+73x+6-2x+2 = x+7, then solve x+8=x+7x+8 = x+7 carefully
Expand to get 3x+6−2x+2=x+73x+6-2x+2 = x+7, then solve x=−1x = -1 and check domains
Expand to get 3x−6−2x−2=x+73x-6-2x-2 = x+7, then solve x=15x = 15 and verify validity
Expand to get 3x+6−2x+2=x+73x+6-2x+2 = x+7, then conclude no solution exists
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Math 3 Quiz

Math 3 Quiz: Solving Rational Equations

Practice Solving Rational Equations in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Rational Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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

To solve 3x−1−2x+2=x+7(x−1)(x+2)\frac{3}{x-1} - \frac{2}{x+2} = \frac{x+7}{(x-1)(x+2)}, a student multiplies both sides by (x−1)(x+2)(x-1)(x+2) and gets 3(x+2)−2(x−1)=x+73(x+2) - 2(x-1) = x+7. What is the next step, and what should the student be careful about?

  1. Expand to get 3x+6−2x+2=x+73x+6-2x+2 = x+7, then solve x+8=x+7x+8 = x+7 carefully
  2. Expand to get 3x+6−2x+2=x+73x+6-2x+2 = x+7, then solve x=−1x = -1 and check domains
  3. Expand to get 3x−6−2x−2=x+73x-6-2x-2 = x+7, then solve x=15x = 15 and verify validity
  4. Expand to get 3x+6−2x+2=x+73x+6-2x+2 = x+7, then conclude no solution exists (correct answer)
Explanation: Expanding 3(x+2)−2(x−1)=x+73(x+2) - 2(x-1) = x+7: 3x+6−2x+2=x+73x+6-2x+2 = x+7, which simplifies to x+8=x+7x+8 = x+7. This gives 8=78 = 7, which is impossible. Therefore, the equation has no solution. The student should conclude that despite the algebraic manipulation being valid, the resulting contradiction indicates that no value of xx satisfies the original equation. This is different from having an extraneous solution (where a solution exists algebraically but violates domain restrictions) - here, no solution exists at all.

Question 2

A student solving x+1x−3+2x−1x+2=5x+8(x−3)(x+2)\frac{x+1}{x-3} + \frac{2x-1}{x+2} = \frac{5x+8}{(x-3)(x+2)} makes an error and concludes that x=3x = 3 is a solution. What type of error did the student most likely make?

  1. Arithmetic error in expanding products, leading to an incorrect coefficient
  2. Failed to check that x=3x = 3 makes a denominator zero in the original equation (correct answer)
  3. Incorrectly combined fractions before clearing denominators completely
  4. Used an invalid method for solving the resulting linear equation
Explanation: In the original equation x+1x−3+2x−1x+2=5x+8(x−3)(x+2)\frac{x+1}{x-3} + \frac{2x-1}{x+2} = \frac{5x+8}{(x-3)(x+2)}, when x=3x = 3, the terms x+1x−3\frac{x+1}{x-3} and 5x+8(x−3)(x+2)\frac{5x+8}{(x-3)(x+2)} have denominators that equal zero (since x−3=0x-3 = 0). This makes these expressions undefined. Even if x=3x = 3 satisfies the algebraic equation obtained after clearing denominators, it cannot be a solution to the original rational equation because it violates the domain restrictions. The student's most likely error was failing to check whether the algebraically obtained solution makes any denominator zero in the original equation.

Question 3

Solve x2−1x+2=x−3+5x+2\frac{x^2-1}{x+2} = x-3 + \frac{5}{x+2}. How many solutions does this equation have?

  1. One solution, because one potential solution is extraneous (correct answer)
  2. Zero solutions, because both potential solutions are extraneous
  3. Two solutions, because both potential solutions are valid
  4. Infinitely many solutions, because the equation simplifies to an identity
Explanation: When solving rational equations, you must always check for extraneous solutions—apparent solutions that don't actually satisfy the original equation due to restrictions in the domain. Let's solve this step by step. First, subtract 5x+2\frac{5}{x+2} from both sides: x2−1x+2−5x+2=x−3\frac{x^2-1}{x+2} - \frac{5}{x+2} = x-3 Combine the fractions on the left: x2−1−5x+2=x−3\frac{x^2-1-5}{x+2} = x-3 x2−6x+2=x−3\frac{x^2-6}{x+2} = x-3 Multiply both sides by (x+2)(x+2), noting that x≠−2x \neq -2: x2−6=(x−3)(x+2)x^2-6 = (x-3)(x+2) x2−6=x2−x−6x^2-6 = x^2-x-6 Subtract x2−6x^2-6 from both sides: 0=−x0 = -x x=0x = 0 Now check this solution in the original equation. When x=0x = 0: the left side gives 0−10+2=−12\frac{0-1}{0+2} = -\frac{1}{2}, and the right side gives 0−3+50+2=−3+52=−120-3+\frac{5}{0+2} = -3+\frac{5}{2} = -\frac{1}{2}. Since both sides equal −12-\frac{1}{2}, x=0x = 0 is valid. The algebraic manipulation yielded only one potential solution, and it's not extraneous. Therefore, the equation has exactly one solution. Choice B is wrong because we found one valid solution, not zero. Choice C is incorrect because we only found one potential solution, not two. Choice D is wrong because the equation doesn't simplify to an identity—it has a specific solution. Always verify solutions in rational equations by substituting back into the original equation, and remember that the domain excludes values that make denominators zero.

Question 4

Consider the equation 1x−1+1x+1=2xx2−1\frac{1}{x-1} + \frac{1}{x+1} = \frac{2x}{x^2-1}. A student claims this equation has no solution. Is the student correct?

  1. Yes, because multiplying through creates a contradiction with no solutions
  2. Yes, because any solution would make the denominators zero
  3. No, because the equation has exactly one valid solution x=0x = 0
  4. No, because the equation is satisfied by infinitely many values of xx (correct answer)
Explanation: Note that x2−1=(x−1)(x+1)x^2-1 = (x-1)(x+1). Multiplying both sides by (x−1)(x+1)(x-1)(x+1): (x+1)+(x−1)1=2x\frac{(x+1) + (x-1)}{1} = 2x, which simplifies to 2x=2x2x = 2x. This is an identity that's true for all values of xx except those that make the original denominators zero (x≠±1x \neq \pm 1). Therefore, the solution set is all real numbers except x=1x = 1 and x=−1x = -1. The student is incorrect; the equation has infinitely many solutions.

Question 5

A student attempts to solve 3x−2=x+1x2−4+2x+2\frac{3}{x-2} = \frac{x+1}{x^2-4} + \frac{2}{x+2} and finds x=2x = 2 as a potential solution. What should the student conclude?

  1. x=2x = 2 is the correct and only solution to the equation
  2. x=2x = 2 is valid, but there is also a second solution x=−2x = -2
  3. x=2x = 2 is an extraneous solution because it makes denominators zero (correct answer)
  4. x=2x = 2 should be rejected, and the equation has no solutions
Explanation: When solving rational equations, you must always check whether your solutions make any denominator equal to zero, as this would make the equation undefined. Let's examine what happens when x=2x = 2. Looking at the original equation 3x−2=x+1x2−4+2x+2\frac{3}{x-2} = \frac{x+1}{x^2-4} + \frac{2}{x+2}, we need to check each denominator. When x=2x = 2:
  • The first term has denominator x−2=2−2=0x-2 = 2-2 = 0
  • The middle term has denominator x2−4=4−4=0x^2-4 = 4-4 = 0
  • The third term has denominator x+2=2+2=4x+2 = 2+2 = 4 (this is fine)
Since two denominators become zero when x=2x = 2, this value makes the original equation undefined. Therefore, x=2x = 2 is an extraneous solution that must be rejected. Answer choice A is wrong because x=2x = 2 cannot be correct when it makes denominators zero. Answer choice B is incorrect because while x=−2x = -2 would also create undefined denominators (making x+2=0x+2 = 0 and x2−4=0x^2-4 = 0), the main issue is that x=2x = 2 itself is invalid. Answer choice D goes too far—while x=2x = 2 must be rejected, we'd need to solve the equation properly to determine if other valid solutions exist. The key lesson: whenever you solve rational equations, always substitute your solutions back into the original equation to verify they don't make any denominator zero. Any solution that does this is extraneous and must be discarded, regardless of whether it satisfies the algebraic manipulations.

Question 6

The equation 2x−1x2−4−1x−2=3x+2\frac{2x-1}{x^2-4} - \frac{1}{x-2} = \frac{3}{x+2} is solved by first factoring x2−4x^2-4. After clearing denominators and simplifying, what type of equation results?

  1. A linear equation in xx with one potential solution to check for validity (correct answer)
  2. A quadratic equation in xx with two potential solutions to check for validity
  3. A contradiction (like 0=50 = 5) indicating no solutions exist
  4. An identity (like 0=00 = 0) indicating infinitely many solutions in the domain
Explanation: First, factor x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2). The equation becomes 2x−1(x−2)(x+2)−1x−2=3x+2\frac{2x-1}{(x-2)(x+2)} - \frac{1}{x-2} = \frac{3}{x+2}. The domain excludes x=2x = 2 and x=−2x = -2. Multiplying through by (x−2)(x+2)(x-2)(x+2): (2x−1)−(x+2)=3(x−2)(2x-1) - (x+2) = 3(x-2). Simplifying the left side: 2x−1−x−2=x−32x-1-x-2 = x-3. The right side: 3x−63x-6. So we have x−3=3x−6x-3 = 3x-6, which gives −3=2x-3 = 2x, so x=−32x = -\frac{3}{2}. This is a linear equation with one solution. Since x=−32x = -\frac{3}{2} doesn't equal 2 or -2, it doesn't violate domain restrictions. Checking: substitute x=−32x = -\frac{3}{2} into the original equation to verify it's valid.