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College Algebra Quiz

College Algebra Quiz: Solving Linear Equations Including Fractions

Practice Solving Linear Equations Including Fractions 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 / 20

0 of 20 answered

Cost sharing: solve 14x+18=30\frac{1}{4}x+18=3041​x+18=30 dollars for xxx.

Select an answer to continue

What this quiz covers

This quiz focuses on Solving Linear Equations Including Fractions, 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

Cost sharing: solve 14x+18=30\frac{1}{4}x+18=3041​x+18=30 dollars for xxx.

  1. x=48x=48x=48 (correct answer)
  2. x=12x=12x=12
  3. x=72x=72x=72
  4. x=36x=36x=36

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation \frac{1}{4}x + 18 = 30 involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice A, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution x=48. A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as forgetting to multiply by the reciprocal. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 2

Mixing: solve x20+14=25\frac{x}{20}+\frac{1}{4}=\frac{2}{5}20x​+41​=52​ for xxx mL.

  1. x=1x=1x=1
  2. x=2x=2x=2
  3. x=3x=3x=3 (correct answer)
  4. x=4x=4x=4

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{20} + \frac{1}{4} = \frac{2}{5}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=3). A common error, as seen in Choice D, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 3

Dividing quantity: solve x7−27=37\frac{x}{7}-\frac{2}{7}=\frac{3}{7}7x​−72​=73​ for xxx.

  1. x=1x=1x=1
  2. x=3x=3x=3
  3. x=5x=5x=5 (correct answer)
  4. x=7x=7x=7

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{7} - \frac{2}{7} = \frac{3}{7}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=5). A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 4

Recipe scaling: solve x3+14=1112\frac{x}{3}+\frac{1}{4}=\frac{11}{12}3x​+41​=1211​ for xxx cups.

  1. x=1x=1x=1
  2. x=2x=2x=2 (correct answer)
  3. x=32x=\frac{3}{2}x=23​
  4. x=52x=\frac{5}{2}x=25​

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{3} + \frac{1}{4} = \frac{11}{12}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=2). A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 5

In cost sharing, solve x4+38=78\frac{x}{4}+\frac{3}{8}=\frac{7}{8}4x​+83​=87​ dollars for xxx.

  1. x=1x=1x=1
  2. x=2x=2x=2 (correct answer)
  3. x=12x=\frac{1}{2}x=21​
  4. x=52x=\frac{5}{2}x=25​

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{4} + \frac{3}{8} = \frac{7}{8}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=2). A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 6

Mixing: solve x6+12=56\frac{x}{6}+\frac{1}{2}=\frac{5}{6}6x​+21​=65​ for xxx mL.

  1. x=1x=1x=1
  2. x=2x=2x=2 (correct answer)
  3. x=3x=3x=3
  4. x=4x=4x=4

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{6} + \frac{1}{2} = \frac{5}{6}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=2). A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 7

Which equation is equivalent to 0.25x+13=0.75x−230.25x + \frac{1}{3} = 0.75x - \frac{2}{3}0.25x+31​=0.75x−32​ when all terms are expressed with integer coefficients?

  1. 3x+4=9x−83x + 4 = 9x - 83x+4=9x−8 (correct answer)
  2. 25x+33=75x−6725x + 33 = 75x - 6725x+33=75x−67
  3. x+4=3x−8x + 4 = 3x - 8x+4=3x−8
  4. 75x+100=225x−20075x + 100 = 225x - 20075x+100=225x−200

Explanation: Convert decimals to fractions: 14x+13=34x−23\frac{1}{4}x + \frac{1}{3} = \frac{3}{4}x - \frac{2}{3}41​x+31​=43​x−32​. To clear denominators, multiply by LCD = 12: 12⋅14x+12⋅13=12⋅34x−12⋅2312 \cdot \frac{1}{4}x + 12 \cdot \frac{1}{3} = 12 \cdot \frac{3}{4}x - 12 \cdot \frac{2}{3}12⋅41​x+12⋅31​=12⋅43​x−12⋅32​. This gives 3x+4=9x−83x + 4 = 9x - 83x+4=9x−8. Choice B uses 100 as multiplier unnecessarily. Choice C incorrectly multiplies the xxx terms. Choice D uses 300 as multiplier and makes coefficient errors.

Question 8

A linear equation in the form px+qr=sx+tu+v\frac{px+q}{r} = \frac{sx+t}{u} + vrpx+q​=usx+t​+v is solved by first finding a common denominator. If the LCD is 30, and after clearing denominators the equation becomes 6px+6q=5sx+5t+30v6px + 6q = 5sx + 5t + 30v6px+6q=5sx+5t+30v, what are the possible values for the pair (r,u)(r,u)(r,u)?

  1. (r,u)(r,u)(r,u) could be (5,6)(5,6)(5,6) or (10,15)(10,15)(10,15), among other possibilities (correct answer)
  2. (r,u)(r,u)(r,u) could be (6,5)(6,5)(6,5) or (15,10)(15,10)(15,10), among other possibilities
  3. (r,u)(r,u)(r,u) must be exactly (5,6)(5,6)(5,6) since those are the only divisors that work
  4. (r,u)(r,u)(r,u) could be (3,10)(3,10)(3,10) or (6,15)(6,15)(6,15), among other possibilities

Explanation: When clearing denominators with LCD = 30, we multiply each term by 30denominator\frac{30}{\text{denominator}}denominator30​. The first fraction gets multiplied by 30r\frac{30}{r}r30​, giving coefficient 6 for the pxpxpx term, so 30r=6\frac{30}{r} = 6r30​=6, thus r=5r = 5r=5. The second fraction gets multiplied by 30u\frac{30}{u}u30​, giving coefficient 5 for the sxsxsx term, so 30u=5\frac{30}{u} = 5u30​=5, thus u=6u = 6u=6. However, any common multiple would work: if LCD were 60, we'd have (r,u)=(10,12)(r,u) = (10,12)(r,u)=(10,12), etc. So (5,6)(5,6)(5,6) and (10,15)(10,15)(10,15) are both possible.

Question 9

Cost sharing: solve x3+19=49\frac{x}{3}+\frac{1}{9}=\frac{4}{9}3x​+91​=94​ dollars for xxx.

  1. x=13x=\frac{1}{3}x=31​
  2. x=23x=\frac{2}{3}x=32​
  3. x=1x=1x=1 (correct answer)
  4. x=43x=\frac{4}{3}x=34​

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{3} + \frac{1}{9} = \frac{4}{9}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=1). A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 10

Dividing items: solve x9−19=23\frac{x}{9}-\frac{1}{9}=\frac{2}{3}9x​−91​=32​ for xxx.

  1. x=5x=5x=5
  2. x=6x=6x=6
  3. x=7x=7x=7 (correct answer)
  4. x=8x=8x=8

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{9} - \frac{1}{9} = \frac{2}{3}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=7). A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 11

Recipe: solve x2+13=56\frac{x}{2}+\frac{1}{3}=\frac{5}{6}2x​+31​=65​ for xxx cups.

  1. x=12x=\frac{1}{2}x=21​
  2. x=1x=1x=1 (correct answer)
  3. x=32x=\frac{3}{2}x=23​
  4. x=2x=2x=2

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{2} + \frac{1}{3} = \frac{5}{6}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=1). A common error, as seen in Choice D, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 12

Cost sharing: solve x6+112=512\frac{x}{6}+\frac{1}{12}=\frac{5}{12}6x​+121​=125​ dollars for xxx.

  1. x=1x=1x=1
  2. x=2x=2x=2 (correct answer)
  3. x=3x=3x=3
  4. x=4x=4x=4

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{6} + \frac{1}{12} = \frac{5}{12}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=2). A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 13

A student is solving the equation 2x−13−x+46=x−22\frac{2x-1}{3} - \frac{x+4}{6} = \frac{x-2}{2}32x−1​−6x+4​=2x−2​. After finding a common denominator and combining fractions, what is the resulting equation before solving for xxx?

  1. 4x−2−x−4=3x−64x - 2 - x - 4 = 3x - 64x−2−x−4=3x−6
  2. 2(2x−1)−(x+4)=3(x−2)2(2x-1) - (x+4) = 3(x-2)2(2x−1)−(x+4)=3(x−2) (correct answer)
  3. 4x−2−x−46=3x−66\frac{4x-2-x-4}{6} = \frac{3x-6}{6}64x−2−x−4​=63x−6​
  4. 2x−1−x+46=x−22\frac{2x-1-x+4}{6} = \frac{x-2}{2}62x−1−x+4​=2x−2​

Explanation: To solve this equation, we need a common denominator of 6. Multiplying each fraction: 2(2x−1)6−x+46=3(x−2)6\frac{2(2x-1)}{6} - \frac{x+4}{6} = \frac{3(x-2)}{6}62(2x−1)​−6x+4​=63(x−2)​. Multiplying both sides by 6 gives us 2(2x−1)−(x+4)=3(x−2)2(2x-1) - (x+4) = 3(x-2)2(2x−1)−(x+4)=3(x−2). Choice A incorrectly distributes without maintaining the structure. Choice C shows the fractions before clearing denominators. Choice D uses an incorrect common denominator approach.

Question 14

Recipe scaling: solve x3−16=12\frac{x}{3}-\frac{1}{6}=\frac{1}{2}3x​−61​=21​ for xxx cups.

  1. x=32x=\frac{3}{2}x=23​
  2. x=2x=2x=2 (correct answer)
  3. x=52x=\frac{5}{2}x=25​
  4. x=3x=3x=3

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{3} - \frac{1}{6} = \frac{1}{2}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=2). A common error, as seen in Choice A, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 15

Recipe: solve x5+12=910\frac{x}{5}+\frac{1}{2}=\frac{9}{10}5x​+21​=109​ for xxx cups.

  1. x=1x=1x=1
  2. x=2x=2x=2 (correct answer)
  3. x=3x=3x=3
  4. x=4x=4x=4

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{5} + \frac{1}{2} = \frac{9}{10}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=2). A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 16

Divide snacks: solve x6+13=56\frac{x}{6}+\frac{1}{3}=\frac{5}{6}6x​+31​=65​ for xxx.

  1. x=1x=1x=1
  2. x=2x=2x=2
  3. x=3x=3x=3 (correct answer)
  4. x=4x=4x=4

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{6} + \frac{1}{3} = \frac{5}{6}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=3). A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 17

Mixing: solve x9+23=89\frac{x}{9}+\frac{2}{3}=\frac{8}{9}9x​+32​=98​ for xxx mL.

  1. x=1x=1x=1
  2. x=2x=2x=2 (correct answer)
  3. x=3x=3x=3
  4. x=4x=4x=4

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{9} + \frac{2}{3} = \frac{8}{9}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=2). A common error, as seen in Choice C, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 18

Distance-speed: solve x9−13=16\frac{x}{9}-\frac{1}{3}=\frac{1}{6}9x​−31​=61​ for xxx miles.

  1. x=32x=\frac{3}{2}x=23​
  2. x=3x=3x=3
  3. x=92x=\frac{9}{2}x=29​ (correct answer)
  4. x=6x=6x=6

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{9} - \frac{1}{3} = \frac{1}{6}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=\frac{9}{2}). A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 19

Cost sharing: solve 56x−10=40\frac{5}{6}x-10=4065​x−10=40 dollars for xxx.

  1. x=36x=36x=36
  2. x=60x=60x=60 (correct answer)
  3. x=50x=50x=50
  4. x=72x=72x=72

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation \frac{5}{6}x - 10 = 40 involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice B, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution x=60. A common error, as seen in Choice A, occurs when students misapply fraction arithmetic, such as forgetting to multiply by the reciprocal. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.

Question 20

Dividing supplies: solve x8+14=58\frac{x}{8}+\frac{1}{4}=\frac{5}{8}8x​+41​=85​ for xxx.

  1. x=1x=1x=1
  2. x=2x=2x=2
  3. x=3x=3x=3 (correct answer)
  4. x=4x=4x=4

Explanation: This question tests solving linear equations including fractions, a foundational skill in college algebra. Solving such equations involves isolating the variable by applying inverse operations and managing fractions correctly, which often requires finding common denominators and simplifying. For the given problem, the equation (\frac{x}{8} + \frac{1}{4} = \frac{5}{8}) involves fractions that must be carefully managed to isolate the variable. The correct answer, Choice C, results from correctly applying fraction addition/subtraction and inverse operations, leading to the solution (x=3). A common error, as seen in Choice B, occurs when students misapply fraction arithmetic, such as incorrectly finding a common denominator. To support student learning, emphasize the importance of checking work for arithmetic errors, using clear fraction manipulation techniques, and verifying solutions within the context of the problem. Encourage practice with a variety of fraction-based equations to build confidence.