ACCUPLACER Advanced Algebra & Functions Quiz: Solving Rational Equations
20 questions · exam conditions
0:00
Solving Rational EquationsQuestion 1 of 20

What is the solution set for the rational equation xx−3−3x+1=12x2−2x−3\frac{x}{x-3} - \frac{3}{x+1} = \frac{12}{x^2-2x-3}?

x=−1x = -1
x=3x = 3
{−1,3}\{-1, 3\}
No solution
← Back to quizzes

ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Rational Equations

Practice Solving Rational Equations in ACCUPLACER Advanced Algebra & Functions 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 ACCUPLACER Advanced Algebra & Functions.

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

What is the solution set for the rational equation xx−3−3x+1=12x2−2x−3\frac{x}{x-3} - \frac{3}{x+1} = \frac{12}{x^2-2x-3}?

  1. x=−1x = -1
  2. x=3x = 3
  3. {−1,3}\{-1, 3\}
  4. No solution (correct answer)
Explanation: To solve the equation, first factor the denominator of the term on the right: x2−2x−3=(x−3)(x+1)x^2-2x-3 = (x-3)(x+1). The least common denominator (LCD) is (x−3)(x+1)(x-3)(x+1). The excluded values are x=3x=3 and x=−1x=-1, as these would make the denominators zero. Multiply each term by the LCD: x(x+1)−3(x−3)=12x(x+1) - 3(x-3) = 12. Distribute and simplify: x2+x−3x+9=12⇒x2−2x+9=12⇒x2−2x−3=0x^2+x-3x+9 = 12 \Rightarrow x^2-2x+9 = 12 \Rightarrow x^2-2x-3=0. Factor the resulting quadratic equation: (x−3)(x+1)=0(x-3)(x+1)=0. The potential solutions are x=3x=3 and x=−1x=-1. However, both of these values are excluded values. Therefore, they are extraneous solutions, and the equation has no solution.

Question 2

What is the sum of the solutions to the equation 2x+3x+2=1\frac{2}{x} + \frac{3}{x+2} = 1?

  1. -1
  2. 1
  3. 3 (correct answer)
  4. 5
Explanation: The excluded values are x=0x=0 and x=−2x=-2. The least common denominator is x(x+2)x(x+2). Multiply the entire equation by the LCD: 2(x+2)+3x=1x(x+2)2(x+2) + 3x = 1x(x+2). Simplify the equation: 2x+4+3x=x2+2x⇒5x+4=x2+2x2x+4+3x = x^2+2x \Rightarrow 5x+4 = x^2+2x. Rearrange into a standard quadratic form: x2−3x−4=0x^2-3x-4=0. Factor the quadratic: (x−4)(x+1)=0(x-4)(x+1)=0. The solutions are x=4x=4 and x=−1x=-1. Neither solution is an excluded value, so both are valid. The sum of the solutions is 4+(−1)=34 + (-1) = 3.

Question 3

Solve for xx: x2x−4−8=16x−4\frac{x^2}{x-4} - 8 = \frac{16}{x-4}

  1. x=−4x = -4
  2. x=4x = 4
  3. {−4,4}\{-4, 4\}
  4. No solution (correct answer)
Explanation: First, note that x=4x=4 is an excluded value. To solve, multiply the entire equation by the denominator (x−4)(x-4) to clear the fractions: x2−8(x−4)=16x^2 - 8(x-4) = 16. This simplifies to x2−8x+32=16x^2 - 8x + 32 = 16, which gives x2−8x+16=0x^2 - 8x + 16 = 0. Factoring this gives (x−4)2=0(x-4)^2 = 0, so the only potential solution is x=4x=4. Alternatively, one could first combine the rational expressions: x2−16x−4=8\frac{x^2-16}{x-4} = 8. Factoring the numerator gives (x−4)(x+4)x−4=8\frac{(x-4)(x+4)}{x-4} = 8, which simplifies to x+4=8x+4=8, yielding x=4x=4. In both methods, the potential solution x=4x=4 is an excluded value, making it an extraneous solution. Therefore, the equation has no solution.

Question 4

What is the solution set for the equation x−7x−7=1\frac{x-7}{x-7} = 1?

  1. No solution
  2. x=7x = 7
  3. The set of all real numbers
  4. The set of all real numbers except 7 (correct answer)
Explanation: The expression x−7x−7\frac{x-7}{x-7} is equal to 1 for any value of xx for which the expression is defined. The expression is undefined when the denominator is zero, which occurs when x−7=0x-7=0, or x=7x=7. For all other real numbers, the equation simplifies to 1=11=1, which is always true. Therefore, the solution set is all real numbers except for the excluded value, 7.

Question 5

The solution to 2xx−4−8x−4=1\frac{2x}{x-4} - \frac{8}{x-4} = 1 is which of the following?

  1. x=4x = 4
  2. x=6x = 6
  3. The set of all real numbers
  4. No solution (correct answer)
Explanation: First, combine the terms on the left side since they have a common denominator: 2x−8x−4=1\frac{2x-8}{x-4} = 1. The excluded value is x=4x=4. Next, factor the numerator: 2(x−4)x−4=1\frac{2(x-4)}{x-4} = 1. For any value of xx other than 4, the (x−4)(x-4) terms cancel, leaving the equation 2=12=1. This is a contradiction. Since this statement is false, there is no value of xx for which the original equation is true. Therefore, there is no solution.

Question 6

What is the product of the solutions to the equation 2x−2+3x−1−2=02x^{-2} + 3x^{-1} - 2 = 0?

  1. -2
  2. -1 (correct answer)
  3. −12-\frac{1}{2}
  4. −14-\frac{1}{4}
Explanation: The equation can be rewritten using positive exponents as 2x2+3x−2=0\frac{2}{x^2} + \frac{3}{x} - 2 = 0. The excluded value is x=0x=0. Let u=1xu = \frac{1}{x}. The equation becomes a quadratic in terms of uu: 2u2+3u−2=02u^2 + 3u - 2 = 0. Factoring this gives (2u−1)(u+2)=0(2u-1)(u+2) = 0. So, u=12u = \frac{1}{2} or u=−2u = -2. Substitute back 1x\frac{1}{x} for uu. If 1x=12\frac{1}{x} = \frac{1}{2}, then x=2x=2. If 1x=−2\frac{1}{x} = -2, then x=−12x = -\frac{1}{2}. Both solutions are valid. The product of the solutions is 2×(−12)=−12 \times (-\frac{1}{2}) = -1.

Question 7

For what value(s) of xx does x+1x−3−2x−3=13\frac{x+1}{x-3} - \frac{2}{x-3} = \frac{1}{3}?

  1. x=0x = 0 (correct answer)
  2. x=6x = 6
  3. x=3x = 3
  4. x=−6x = -6
Explanation: First note that x≠3x \neq 3 (excluded value). Combine fractions on the left: x+1−2x−3=x−1x−3=13\frac{x+1-2}{x-3} = \frac{x-1}{x-3} = \frac{1}{3}. Cross-multiply: 3(x−1)=1(x−3)3(x-1) = 1(x-3), giving 3x−3=x−33x - 3 = x - 3, so 2x=02x = 0 and x=0x = 0. Since x=0x = 0 is not excluded, it's valid. Choice B results from the error 3x−3=x+33x - 3 = x + 3. Choice C is the excluded value. Choice D comes from sign errors in cross-multiplication.

Question 8

For which value of xx is the equation x2x−2=4x−2\frac{x^2}{x-2} = \frac{4}{x-2} satisfied?

  1. x=2x = 2 only
  2. x=−2x = -2 only (correct answer)
  3. x=2x = 2 and x=−2x = -2
  4. x=−2x = -2 and x=0x = 0
Explanation: First note that x=2x = 2 is excluded since it makes the denominator zero. Since both sides have the same denominator (x−2)(x-2), multiply both sides by (x−2)(x-2): x2=4x^2 = 4. This gives x=2x = 2 or x=−2x = -2. However, x=2x = 2 is excluded, so only x=−2x = -2 is valid. Choice A gives the excluded value only. Choice C includes the excluded value. Choice D incorrectly includes x=0x = 0, which doesn't satisfy x2=4x^2 = 4.

Question 9

Which statement is true about the equation 2xx+1=2x+2x+1\frac{2x}{x+1} = \frac{2x+2}{x+1}?

  1. The solution is x=−1x = -1 only
  2. The solution is x=−1x = -1 and x=1x = 1
  3. The solution is x=1x = 1 only
  4. There is no solution to this equation (correct answer)
Explanation: First identify that x=−1x = -1 is excluded. Since both sides have the same denominator (x+1)(x+1), we can multiply both sides by (x+1)(x+1) to get 2x=2x+22x = 2x + 2. This simplifies to 0=20 = 2, which is impossible. Therefore, no solution exists. Choice A uses the excluded value. Choice B includes both the excluded value and an incorrect solution. Choice C comes from incorrectly solving 2x=2x+22x = 2x + 2 as x=1x = 1.

Question 10

What is the solution to the equation xx−5+1x−1=20x2−6x+5\frac{x}{x-5} + \frac{1}{x-1} = \frac{20}{x^2-6x+5}?

  1. x=−5x = -5 (correct answer)
  2. x=5x = 5
  3. {−5,5}\{-5, 5\}
  4. No solution
Explanation: Factor the denominator on the right side: x2−6x+5=(x−5)(x−1)x^2-6x+5 = (x-5)(x-1). The excluded values are x=5x=5 and x=1x=1. Multiply the entire equation by the LCD, (x−5)(x−1)(x-5)(x-1): x(x−1)+1(x−5)=20x(x-1) + 1(x-5) = 20. Distribute and simplify: x2−x+x−5=20⇒x2−5=20x^2-x+x-5=20 \Rightarrow x^2-5=20. This leads to x2=25x^2=25, so the potential solutions are x=5x=5 and x=−5x=-5. We must check these against the excluded values. The potential solution x=5x=5 is an excluded value, so it is extraneous. The potential solution x=−5x=-5 is not an excluded value, so it is the only valid solution.

Question 11

If 1R=1R1+1R2\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}, which of the following is an expression for RR in terms of R1R_1 and R2R_2?

  1. R=R1+R2R = R_1 + R_2
  2. R=R1+R2R1R2R = \frac{R_1+R_2}{R_1R_2}
  3. R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2} (correct answer)
  4. R=R1R2R = R_1R_2
Explanation: To solve for RR, we first combine the fractions on the right side. The least common denominator for R1R_1 and R2R_2 is R1R2R_1R_2. So, 1R1+1R2=R2R1R2+R1R1R2=R1+R2R1R2\frac{1}{R_1} + \frac{1}{R_2} = \frac{R_2}{R_1R_2} + \frac{R_1}{R_1R_2} = \frac{R_1+R_2}{R_1R_2}. The equation becomes 1R=R1+R2R1R2\frac{1}{R} = \frac{R_1+R_2}{R_1R_2}. To solve for RR, we can take the reciprocal of both sides, which gives R=R1R2R1+R2R = \frac{R_1R_2}{R_1+R_2}.

Question 12

A student is solving a rational equation and correctly simplifies it to x−5=x+1x-5 = x+1. What can the student conclude?

  1. The solution is x=0x=0.
  2. The solution is the set of all real numbers.
  3. There is no solution to the equation. (correct answer)
  4. The solution depends on the excluded values.
Explanation: The simplified equation is x−5=x+1x-5 = x+1. If we subtract xx from both sides, we get −5=1-5 = 1. This is a contradiction, a statement that is never true. When the process of solving an equation leads to a contradiction, it means there is no value of the variable that can make the original equation true. Therefore, the equation has no solution. The excluded values are irrelevant at this stage, as no potential solutions were even found.

Question 13

Find the solution for xx in the equation 2xx+3−1x−3=2\frac{2x}{x+3} - \frac{1}{x-3} = 2.

  1. x=−3x = -3
  2. x=157x = \frac{15}{7} (correct answer)
  3. x=3x = 3
  4. x=215x = \frac{21}{5}
Explanation: The excluded values are x=−3x=-3 and x=3x=3. The LCD is (x+3)(x−3)(x+3)(x-3). Multiply the equation by the LCD: 2x(x−3)−1(x+3)=2(x+3)(x−3)2x(x-3) - 1(x+3) = 2(x+3)(x-3). Distribute on both sides: 2x2−6x−x−3=2(x2−9)2x^2 - 6x - x - 3 = 2(x^2 - 9). Simplify: 2x2−7x−3=2x2−182x^2 - 7x - 3 = 2x^2 - 18. The 2x22x^2 terms cancel out, leaving a linear equation: −7x−3=−18-7x - 3 = -18. Add 3 to both sides: −7x=−15-7x = -15. Divide by -7: x=157x = \frac{15}{7}. This solution is not an excluded value, so it is valid.

Question 14

What is the sum of the solutions to the equation 52x−1=x3\frac{5}{2x-1} = \frac{x}{3}?

  1. −52-\frac{5}{2}
  2. 12\frac{1}{2} (correct answer)
  3. 3
  4. 72\frac{7}{2}
Explanation: The excluded value is 2x−1=0⇒x=122x-1=0 \Rightarrow x=\frac{1}{2}. To solve, cross-multiply: 5(3)=x(2x−1)5(3) = x(2x-1). This gives 15=2x2−x15 = 2x^2-x. Rearrange into standard quadratic form: 2x2−x−15=02x^2-x-15=0. This can be factored as (2x+5)(x−3)=0(2x+5)(x-3)=0. The solutions are x=−52x=-\frac{5}{2} and x=3x=3. Neither is the excluded value. The sum of the solutions is −52+3=−52+62=12-\frac{5}{2} + 3 = -\frac{5}{2} + \frac{6}{2} = \frac{1}{2}.

Question 15

Which of the following describes the first two steps to solve the equation xx+1−2x2−1=3\frac{x}{x+1} - \frac{2}{x^2-1} = 3?

  1. Factor x2−1x^2-1, then subtract 3 from both sides.
  2. Factor x2−1x^2-1, then multiply all terms by the least common denominator. (correct answer)
  3. Multiply all terms by (x+1)(x2−1)(x+1)(x^2-1), then simplify the resulting expression.
  4. Identify the excluded value as x=−1x=-1, then cross-multiply the terms.
Explanation: The standard procedure for solving rational equations is to first find the least common denominator (LCD) of all terms. To do this, all denominators must be factored. The denominator x2−1x^2-1 can be factored into (x+1)(x−1)(x+1)(x-1). This reveals that the LCD is (x+1)(x−1)(x+1)(x-1). The second step is to multiply every term in the equation by this LCD to eliminate the denominators. This transforms the rational equation into a polynomial equation, which can then be solved.

Question 16

What is the product of the solutions to the equation x+1x=2x+23x−5\frac{x+1}{x} = \frac{2x+2}{3x-5}?

  1. -5 (correct answer)
  2. -1
  3. 4
  4. 5
Explanation: The excluded values are x=0x=0 and x=5/3x=5/3. We can solve by cross-multiplication: (x+1)(3x−5)=x(2x+2)(x+1)(3x-5) = x(2x+2). Expand both sides: 3x2−5x+3x−5=2x2+2x3x^2 - 5x + 3x - 5 = 2x^2 + 2x. Combine like terms: 3x2−2x−5=2x2+2x3x^2 - 2x - 5 = 2x^2 + 2x. Move all terms to one side to form a quadratic equation: x2−4x−5=0x^2 - 4x - 5 = 0. Factor the quadratic: (x−5)(x+1)=0(x-5)(x+1)=0. The solutions are x=5x=5 and x=−1x=-1. Neither is an excluded value. The product of the solutions is 5×(−1)=−55 \times (-1) = -5.

Question 17

What is the solution set for the equation x2−x+4x−1=−1\frac{x^2-x+4}{x-1} = -1?

  1. x=−1x = -1
  2. x=3x = 3
  3. {−1,3}\{-1, 3\}
  4. No solution (correct answer)
Explanation: The excluded value is x=1x=1. To solve, multiply both sides by the denominator (x−1)(x-1): x2−x+4=−1(x−1)x^2-x+4 = -1(x-1). Distribute on the right side: x2−x+4=−x+1x^2-x+4 = -x+1. Add xx to both sides: x2+4=1x^2+4=1. Subtract 4 from both sides: x2=−3x^2 = -3. There is no real number that can be squared to produce a negative result. Therefore, the equation has no real solutions.

Question 18

The equation xx−2−1x=4x(x−2)\frac{x}{x-2} - \frac{1}{x} = \frac{4}{x(x-2)} has an extraneous solution. What is the value of this extraneous solution?

  1. 0
  2. 2 (correct answer)
  3. 4
  4. The equation has no extraneous solutions.
Explanation: An extraneous solution is a value obtained during the solving process that is not a valid solution because it is an excluded value (i.e., it makes a denominator zero). The denominators are xx and x−2x-2, so the excluded values are x=0x=0 and x=2x=2. To solve, multiply by the LCD, x(x−2)x(x-2): x(x)−1(x−2)=4x(x) - 1(x-2) = 4. Simplify: x2−x+2=4⇒x2−x−2=0x^2-x+2=4 \Rightarrow x^2-x-2=0. Factor the quadratic: (x−2)(x+1)=0(x-2)(x+1)=0. The potential solutions are x=2x=2 and x=−1x=-1. The value x=−1x=-1 is a valid solution. The value x=2x=2 is an excluded value, so it is the extraneous solution.

Question 19

What is the sum of the excluded values for the equation 2xx2−16−3x2−2x−8=5\frac{2x}{x^2-16} - \frac{3}{x^2-2x-8} = 5?

  1. -2 (correct answer)
  2. 0
  3. 2
  4. 6
Explanation: The excluded values are the values of xx that make any denominator zero. We need to find the roots of each denominator. For the first denominator, x2−16=0⇒(x−4)(x+4)=0x^2-16=0 \Rightarrow (x-4)(x+4)=0, so x=4x=4 and x=−4x=-4 are excluded. For the second denominator, x2−2x−8=0⇒(x−4)(x+2)=0x^2-2x-8=0 \Rightarrow (x-4)(x+2)=0, so x=4x=4 and x=−2x=-2 are excluded. The set of unique excluded values is {−4,−2,4}\{-4, -2, 4\}. The sum of these values is −4+(−2)+4=−2-4 + (-2) + 4 = -2.

Question 20

For which of the following values of xx is the equation x+5x2−4−2x2+3x−10=1x+2\frac{x+5}{x^2-4} - \frac{2}{x^2+3x-10} = \frac{1}{x+2} undefined?

  1. {−2,2}\{-2, 2\}
  2. {−5,2}\{-5, 2\}
  3. {−5,−2,2}\{-5, -2, 2\} (correct answer)
  4. {−5,−2,2,5}\{-5, -2, 2, 5\}
Explanation: An equation is undefined for any value of xx that makes any denominator equal to zero. We must find the roots of all three denominators. First denominator: x2−4=(x−2)(x+2)x^2-4 = (x-2)(x+2), so x=2x=2 and x=−2x=-2 are excluded. Second denominator: x2+3x−10=(x+5)(x−2)x^2+3x-10 = (x+5)(x-2), so x=−5x=-5 and x=2x=2 are excluded. Third denominator: x+2x+2, so x=−2x=-2 is excluded. The complete set of unique excluded values is the union of these sets: {−5,−2,2}\{-5, -2, 2\}.