ACCUPLACER Advanced Algebra & Functions Quiz: Solving Linear Equations
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Solving Linear EquationsQuestion 1 of 20

If a(x−c)=b(x+d)a(x - c) = b(x + d) for a≠ba \neq b, which of the following gives xx in terms of a,b,c,a, b, c, and dd?

x=ac+bda−bx = \frac{ac + bd}{a - b}
x=ac+bdb−ax = \frac{ac + bd}{b - a}
x=ac−bda+bx = \frac{ac - bd}{a + b}
x=d−ca−bx = \frac{d - c}{a - b}
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Solving Linear Equations

Practice Solving Linear 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 Linear 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

If a(x−c)=b(x+d)a(x - c) = b(x + d) for a≠ba \neq b, which of the following gives xx in terms of a,b,c,a, b, c, and dd?

  1. x=ac+bda−bx = \frac{ac + bd}{a - b} (correct answer)
  2. x=ac+bdb−ax = \frac{ac + bd}{b - a}
  3. x=ac−bda+bx = \frac{ac - bd}{a + b}
  4. x=d−ca−bx = \frac{d - c}{a - b}
Explanation: First, distribute aa and bb on both sides of the equation: ax−ac=bx+bdax - ac = bx + bd. Next, collect all terms with xx on one side and all constant terms on the other side: ax−bx=ac+bdax - bx = ac + bd. Factor out xx from the terms on the left side: x(a−b)=ac+bdx(a - b) = ac + bd. Finally, divide by (a−b)(a - b) to isolate xx: x=ac+bda−bx = \frac{ac + bd}{a - b}.

Question 2

For what value of cc does the equation 3(x+c)−5x=2(x−1)3(x + c) - 5x = 2(x - 1) have a solution of x=12x = \frac{1}{2}?

  1. −23-\frac{2}{3}
  2. 0 (correct answer)
  3. 13\frac{1}{3}
  4. 1
Explanation: Substitute the given solution x=12x = \frac{1}{2} into the equation and then solve for cc. The equation becomes 3(12+c)−5(12)=2(12−1)3(\frac{1}{2} + c) - 5(\frac{1}{2}) = 2(\frac{1}{2} - 1). Simplify each part: 32+3c−52=2(−12)\frac{3}{2} + 3c - \frac{5}{2} = 2(-\frac{1}{2}). Combine the constant terms on the left and simplify the right: −22+3c=−1-\frac{2}{2} + 3c = -1, which simplifies to −1+3c=−1-1 + 3c = -1. Add 1 to both sides: 3c=03c = 0. Finally, divide by 3 to find c=0c = 0.

Question 3

For what value of kk does the equation k(x−2)=3x+7k(x - 2) = 3x + 7 have no solution?

  1. −72-\frac{7}{2}
  2. -3
  3. 0
  4. 3 (correct answer)
Explanation: First, distribute kk on the left side: kx−2k=3x+7kx - 2k = 3x + 7. To analyze the number of solutions, group the xx terms on one side and the constant terms on the other: kx−3x=2k+7kx - 3x = 2k + 7. Factor out xx: (k−3)x=2k+7(k - 3)x = 2k + 7. A linear equation has no solution when the coefficient of xx is zero and the constant term on the other side is non-zero. Set the coefficient of xx to zero: k−3=0k - 3 = 0, which gives k=3k = 3. Now, check if the right side is non-zero for this value of kk: 2(3)+7=6+7=132(3) + 7 = 6 + 7 = 13. Since 13≠013 \neq 0, the equation becomes 0x=130x = 13, which has no solution. Thus, the value is k=3k = 3.

Question 4

What is the solution to the equation 34(x−2)=13x+1\frac{3}{4}(x - 2) = \frac{1}{3}x + 1?

  1. x=185x = \frac{18}{5}
  2. x=195x = \frac{19}{5}
  3. x=6x = 6 (correct answer)
  4. x=365x = \frac{36}{5}
Explanation: To solve the equation, first distribute the 34\frac{3}{4} to get 34x−32=13x+1\frac{3}{4}x - \frac{3}{2} = \frac{1}{3}x + 1. To eliminate the fractions, multiply the entire equation by the least common denominator, which is 12. This yields 12(34x)−12(32)=12(13x)+12(1)12(\frac{3}{4}x) - 12(\frac{3}{2}) = 12(\frac{1}{3}x) + 12(1), which simplifies to 9x−18=4x+129x - 18 = 4x + 12. Next, gather the x-terms on one side and the constants on the other: 9x−4x=12+189x - 4x = 12 + 18, which simplifies to 5x=305x = 30. Finally, divide by 5 to get x=6x = 6.

Question 5

If x3+x4+1=x2\frac{x}{3} + \frac{x}{4} + 1 = \frac{x}{2}, what is the value of 6x+56x + 5?

  1. -67 (correct answer)
  2. -12
  3. -1
  4. 77
Explanation: First, solve for xx. Find the least common denominator of 3, 4, and 2, which is 12. Multiply every term in the equation by 12: 12(x3)+12(x4)+12(1)=12(x2)12(\frac{x}{3}) + 12(\frac{x}{4}) + 12(1) = 12(\frac{x}{2}). This simplifies to 4x+3x+12=6x4x + 3x + 12 = 6x. Combine like terms on the left: 7x+12=6x7x + 12 = 6x. Subtract 7x7x from both sides to get 12=−x12 = -x, which means x=−12x = -12. The question asks for the value of 6x+56x + 5, not xx. Substitute x=−12x = -12 into the expression: 6(−12)+5=−72+5=−676(-12) + 5 = -72 + 5 = -67.

Question 6

When three-fifths of a number is subtracted from 5, the result is equal to one-half the number. What is the number?

  1. 112\frac{11}{2}
  2. 5011\frac{50}{11} (correct answer)
  3. 30
  4. 50
Explanation: Let the number be xx. The phrase 'three-fifths of a number is subtracted from 5' translates to 5−35x5 - \frac{3}{5}x. The phrase 'is equal to one-half the number' translates to =12x= \frac{1}{2}x. So, the equation is 5−35x=12x5 - \frac{3}{5}x = \frac{1}{2}x. To solve for xx, first add 35x\frac{3}{5}x to both sides: 5=12x+35x5 = \frac{1}{2}x + \frac{3}{5}x. To add the fractions on the right, find a common denominator, which is 10: 5=510x+610x5 = \frac{5}{10}x + \frac{6}{10}x. Combine the terms: 5=1110x5 = \frac{11}{10}x. To isolate xx, multiply both sides by the reciprocal of 1110\frac{11}{10}, which is 1011\frac{10}{11}: x=5⋅1011=5011x = 5 \cdot \frac{10}{11} = \frac{50}{11}.

Question 7

If x+ab=x−ba\frac{x + a}{b} = \frac{x - b}{a} and a,b≠0,a≠−ba, b \neq 0, a \neq -b, what is xx in terms of aa and bb?

  1. x=a2+b2a−bx = \frac{a^2 + b^2}{a - b}
  2. x=a2−b2a+bx = \frac{a^2 - b^2}{a + b}
  3. x=−a2+b2a−bx = -\frac{a^2 + b^2}{a - b} (correct answer)
  4. x=b2−a2a+bx = \frac{b^2 - a^2}{a + b}
Explanation: To solve for xx, first cross-multiply: a(x+a)=b(x−b)a(x + a) = b(x - b). Distribute on both sides: ax+a2=bx−b2ax + a^2 = bx - b^2. Collect all terms containing xx on one side and all other terms on the other side: ax−bx=−a2−b2ax - bx = -a^2 - b^2. Factor out xx from the left side: x(a−b)=−(a2+b2)x(a - b) = -(a^2 + b^2). Finally, divide by (a−b)(a - b) to isolate xx: x=−(a2+b2)a−bx = \frac{-(a^2 + b^2)}{a - b}, which can be written as x=−a2+b2a−bx = -\frac{a^2 + b^2}{a - b}.

Question 8

The formula C=59(F−32)C = \frac{5}{9}(F - 32) converts temperatures from Fahrenheit (F) to Celsius (C). Which of the following formulas correctly converts Celsius to Fahrenheit?

  1. F=59C+32F = \frac{5}{9}C + 32
  2. F=95C−32F = \frac{9}{5}C - 32
  3. F=95C+32F = \frac{9}{5}C + 32 (correct answer)
  4. F=95(C+32)F = \frac{9}{5}(C + 32)
Explanation: To solve the equation C=59(F−32)C = \frac{5}{9}(F - 32) for FF, we need to isolate FF. First, multiply both sides by the reciprocal of 59\frac{5}{9}, which is 95\frac{9}{5}. This gives 95C=F−32\frac{9}{5}C = F - 32. Next, to isolate FF, add 32 to both sides of the equation: 95C+32=F\frac{9}{5}C + 32 = F. So, the correct formula is F=95C+32F = \frac{9}{5}C + 32.

Question 9

What is the solution set for 12(4x−6)−x=x−5\frac{1}{2}(4x - 6) - x = x - 5?

  1. No solution (correct answer)
  2. Infinitely many solutions
  3. x=−1x = -1
  4. x=4x = 4
Explanation: First, simplify the left side of the equation. Distribute the 12\frac{1}{2}: 2x−3−x=x−52x - 3 - x = x - 5. Combine the x-terms on the left side: x−3=x−5x - 3 = x - 5. If we try to isolate xx by subtracting xx from both sides, we get −3=−5-3 = -5. This is a false statement, or a contradiction. Because the variable xx has been eliminated and the resulting statement is false, there is no value of xx that can make the original equation true. Therefore, there is no solution.

Question 10

Solve for yy in the equation 0.3(2y−4)=0.5y−0.80.3(2y - 4) = 0.5y - 0.8.

  1. y=2y = 2
  2. y=4y = 4 (correct answer)
  3. y=−2y = -2
  4. y=−4y = -4
Explanation: Expand the left side: 0.3(2y−4)=0.6y−1.20.3(2y - 4) = 0.6y - 1.2. The equation becomes 0.6y−1.2=0.5y−0.80.6y - 1.2 = 0.5y - 0.8. Subtract 0.5y0.5y from both sides: 0.1y−1.2=−0.80.1y - 1.2 = -0.8. Add 1.21.2 to both sides: 0.1y=0.40.1y = 0.4. Divide by 0.10.1: y=4y = 4. Choice A results from incorrectly distributing 0.30.3 as 0.6y−0.120.6y - 0.12 instead of 0.6y−1.20.6y - 1.2. Choice C comes from a sign error when moving terms. Choice D results from confusing the constants during algebraic manipulation.

Question 11

What is the solution to the equation 12[x−3(x−1)]=2x−4\frac{1}{2}[x - 3(x - 1)] = 2x - 4?

  1. 56\frac{5}{6}
  2. 76\frac{7}{6}
  3. 116\frac{11}{6} (correct answer)
  4. 73\frac{7}{3}
Explanation: First, simplify inside the innermost parentheses: 12[x−3x+3]=2x−4\frac{1}{2}[x - 3x + 3] = 2x - 4. Combine like terms inside the brackets: 12[−2x+3]=2x−4\frac{1}{2}[-2x + 3] = 2x - 4. Distribute the 12\frac{1}{2}: −x+32=2x−4-x + \frac{3}{2} = 2x - 4. Add xx to both sides: 32=3x−4\frac{3}{2} = 3x - 4. Add 4 to both sides: 32+4=3x\frac{3}{2} + 4 = 3x. Convert 4 to a fraction with a denominator of 2: 32+82=3x\frac{3}{2} + \frac{8}{2} = 3x, which is 112=3x\frac{11}{2} = 3x. Finally, divide by 3 (or multiply by 13\frac{1}{3}): x=116x = \frac{11}{6}.

Question 12

If 0.2(x−5)+0.3=0.5(x+1)0.2(x - 5) + 0.3 = 0.5(x + 1), what is the value of xx?

  1. -4.9
  2. -4 (correct answer)
  3. -1
  4. -0.67
Explanation: To eliminate the decimals, multiply the entire equation by 10. This gives 10[0.2(x−5)]+10[0.3]=10[0.5(x+1)]10[0.2(x - 5)] + 10[0.3] = 10[0.5(x + 1)], which simplifies to 2(x−5)+3=5(x+1)2(x - 5) + 3 = 5(x + 1). Next, distribute on both sides: 2x−10+3=5x+52x - 10 + 3 = 5x + 5. Combine the constant terms on the left: 2x−7=5x+52x - 7 = 5x + 5. Subtract 2x2x from both sides: −7=3x+5-7 = 3x + 5. Subtract 5 from both sides: −12=3x-12 = 3x. Finally, divide by 3: x=−4x = -4.

Question 13

What is the value of kk that satisfies the equation 2k−35−k+12=−3\frac{2k - 3}{5} - \frac{k + 1}{2} = -3?

  1. -8
  2. 19 (correct answer)
  3. 23
  4. 29
Explanation: To clear the denominators, multiply the entire equation by the least common denominator of 5 and 2, which is 10. This gives 10(2k−35)−10(k+12)=10(−3)10(\frac{2k - 3}{5}) - 10(\frac{k + 1}{2}) = 10(-3). The equation simplifies to 2(2k−3)−5(k+1)=−302(2k - 3) - 5(k + 1) = -30. Distribute the 2 and -5: 4k−6−5k−5=−304k - 6 - 5k - 5 = -30. Combine like terms on the left side: −k−11=−30-k - 11 = -30. Add 11 to both sides: −k=−19-k = -19. Finally, multiply by -1 to solve for kk: k=19k = 19.

Question 14

If 23−12(x−1)=14x\frac{2}{3} - \frac{1}{2}(x - 1) = \frac{1}{4}x, what is xx?

  1. x=−2x = -2
  2. x=29x = \frac{2}{9}
  3. x=149x = \frac{14}{9} (correct answer)
  4. x=2x = 2
Explanation: To solve for xx, first clear the fractions by multiplying the entire equation by the least common denominator of 3, 2, and 4, which is 12. The equation becomes 12(23)−12(12(x−1))=12(14x)12(\frac{2}{3}) - 12(\frac{1}{2}(x - 1)) = 12(\frac{1}{4}x). This simplifies to 8−6(x−1)=3x8 - 6(x - 1) = 3x. Distribute the -6 on the left side: 8−6x+6=3x8 - 6x + 6 = 3x. Combine the constant terms: 14−6x=3x14 - 6x = 3x. Add 6x6x to both sides to collect the xx terms: 14=9x14 = 9x. Finally, divide by 9 to find x=149x = \frac{14}{9}.

Question 15

Find the solution to the equation x−2[x−3(x−4)]=6x - 2[x - 3(x - 4)] = 6.

  1. −185-\frac{18}{5}
  2. -10
  3. 6 (correct answer)
  4. 185\frac{18}{5}
Explanation: To solve this equation, simplify from the innermost parentheses outward. First, distribute the -3: x−2[x−3x+12]=6x - 2[x - 3x + 12] = 6. Combine the x-terms inside the brackets: x−2[−2x+12]=6x - 2[-2x + 12] = 6. Next, distribute the -2: x+4x−24=6x + 4x - 24 = 6. Combine the x-terms on the left side: 5x−24=65x - 24 = 6. Add 24 to both sides: 5x=305x = 30. Finally, divide by 5 to find the solution: x=6x = 6.

Question 16

What is the solution for yy in the equation y−y−12=3−y+23y - \frac{y - 1}{2} = 3 - \frac{y + 2}{3}?

  1. −45-\frac{4}{5}
  2. 115\frac{11}{5} (correct answer)
  3. 195\frac{19}{5}
  4. 5
Explanation: To solve this equation, first eliminate the fractions by multiplying every term by the least common denominator of 2 and 3, which is 6. This gives 6(y)−6(y−12)=6(3)−6(y+23)6(y) - 6(\frac{y - 1}{2}) = 6(3) - 6(\frac{y + 2}{3}). Simplify: 6y−3(y−1)=18−2(y+2)6y - 3(y - 1) = 18 - 2(y + 2). Be careful to distribute the negative signs: 6y−3y+3=18−2y−46y - 3y + 3 = 18 - 2y - 4. Combine like terms on each side: 3y+3=14−2y3y + 3 = 14 - 2y. Add 2y2y to both sides: 5y+3=145y + 3 = 14. Subtract 3 from both sides: 5y=115y = 11. Finally, divide by 5: y=115y = \frac{11}{5}.

Question 17

A solution is 20% acid. Another solution is 50% acid. How many liters of the 50% solution must be added to 15 liters of the 20% solution to produce a new solution that is 40% acid?

  1. 10
  2. 15
  3. 25
  4. 30 (correct answer)
Explanation: Let xx be the number of liters of the 50% solution to be added. The total amount of acid in the final mixture is the sum of the amounts of acid from the two initial solutions. The amount of acid is the concentration times the volume. So, the equation is 0.20(15)+0.50(x)=0.40(15+x)0.20(15) + 0.50(x) = 0.40(15 + x). Calculate the known products: 3+0.50x=0.40(15+x)3 + 0.50x = 0.40(15 + x). Distribute on the right side: 3+0.50x=6+0.40x3 + 0.50x = 6 + 0.40x. Subtract 0.40x0.40x from both sides: 3+0.10x=63 + 0.10x = 6. Subtract 3 from both sides: 0.10x=30.10x = 3. Finally, divide by 0.10: x=30.10=30x = \frac{3}{0.10} = 30. So, 30 liters must be added.

Question 18

The perimeter of a rectangle is 60 centimeters. The length is 32\frac{3}{2} of the width. What is the length of the rectangle in centimeters?

  1. 12
  2. 18 (correct answer)
  3. 24
  4. 36
Explanation: The formula for the perimeter of a rectangle is P=2l+2wP = 2l + 2w. We are given P=60P = 60 and l=32wl = \frac{3}{2}w. Substitute these into the formula: 60=2(32w)+2w60 = 2(\frac{3}{2}w) + 2w. Simplify the equation: 60=3w+2w60 = 3w + 2w, which becomes 60=5w60 = 5w. Solve for the width, ww, by dividing by 5: w=12w = 12. The question asks for the length, ll. Substitute the value of ww back into the relationship l=32wl = \frac{3}{2}w: l=32(12)=18l = \frac{3}{2}(12) = 18. The length is 18 centimeters.

Question 19

If 4(x−13)=2−x4(x - \frac{1}{3}) = 2 - x, what is the value of xx?

  1. 25\frac{2}{5}
  2. 715\frac{7}{15}
  3. 23\frac{2}{3} (correct answer)
  4. 109\frac{10}{9}
Explanation: First, distribute the 4 on the left side: 4x−43=2−x4x - \frac{4}{3} = 2 - x. Add xx to both sides to collect the variable terms: 5x−43=25x - \frac{4}{3} = 2. Add 43\frac{4}{3} to both sides to collect the constant terms: 5x=2+435x = 2 + \frac{4}{3}. To add the terms on the right, find a common denominator: 5x=63+435x = \frac{6}{3} + \frac{4}{3}, which simplifies to 5x=1035x = \frac{10}{3}. Finally, divide both sides by 5: x=103÷5=103⋅15=1015=23x = \frac{10}{3} \div 5 = \frac{10}{3} \cdot \frac{1}{5} = \frac{10}{15} = \frac{2}{3}.

Question 20

The sum of two numbers is 36. One-half of the larger number is 3 more than the smaller number. What is the value of the larger number?

  1. 10
  2. 22
  3. 24
  4. 26 (correct answer)
Explanation: Let LL be the larger number and SS be the smaller number. The given information translates to two equations: 1) L+S=36L + S = 36 and 2) 12L=S+3\frac{1}{2}L = S + 3. From equation 1, we can express SS as S=36−LS = 36 - L. Substitute this expression for SS into equation 2: 12L=(36−L)+3\frac{1}{2}L = (36 - L) + 3. Simplify the right side: 12L=39−L\frac{1}{2}L = 39 - L. To solve for LL, add LL to both sides: 12L+L=39\frac{1}{2}L + L = 39, which gives 32L=39\frac{3}{2}L = 39. Finally, multiply both sides by 23\frac{2}{3} to find LL: L=39⋅23=13⋅2=26L = 39 \cdot \frac{2}{3} = 13 \cdot 2 = 26. The larger number is 26.