College Algebra Quiz: Solving Linear Equations Including Fractions

20 questions · exam conditions

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Solving Linear Equations Including FractionsQuestion 1 of 20

Cost sharing: solve $\frac{1}{4}x+18=30$ dollars for $x$ .

x=48

x=12

x=72

x=36

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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.

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 $\frac{1}{4}x+18=30$ dollars for $x$ .

$x=48$ (correct answer)
$x=12$
$x=72$
$x=36$

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

In cost sharing, solve $\frac{x}{4}+\frac{3}{8}=\frac{7}{8}$ dollars for $x$ .

$x=1$
$x=2$ (correct answer)
$x=\frac{1}{2}$
$x=\frac{5}{2}$

\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 3

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

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

Explanation: When clearing denominators with LCD = 30, we multiply each term by

\frac{30}{\text{denominator}}

. The first fraction gets multiplied by

\frac{30}{r}

, giving coefficient 6 for the

px

term, so

\frac{30}{r} = 6

, thus

r = 5

. The second fraction gets multiplied by

\frac{30}{u}

, giving coefficient 5 for the

sx

term, so

\frac{30}{u} = 5

, thus

u = 6

. However, any common multiple would work: if LCD were 60, we'd have

(r,u) = (10,12)

, etc. So

(5,6)

and

(10,15)

are both possible.

Question 4

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

$4x - 2 - x - 4 = 3x - 6$
$2(2x-1) - (x+4) = 3(x-2)$ (correct answer)
$\frac{4x-2-x-4}{6} = \frac{3x-6}{6}$
$\frac{2x-1-x+4}{6} = \frac{x-2}{2}$

Explanation: To solve this equation, we need a common denominator of 6. Multiplying each fraction:

\frac{2(2x-1)}{6} - \frac{x+4}{6} = \frac{3(x-2)}{6}

. Multiplying both sides by 6 gives us

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 5

Recipe: solve $\frac{x}{5}+\frac{1}{2}=\frac{9}{10}$ for $x$ cups.

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

\frac{x}{5} + \frac{1}{2} = \frac{9}{10}

x=2

Question 6

Cost sharing: solve $\frac{x}{5}-\frac{1}{10}=\frac{3}{10}$ dollars for $x$ .

$x=1$
$x=2$ (correct answer)
$x=\frac{3}{2}$
$x=\frac{5}{2}$

\frac{x}{5} - \frac{1}{10} = \frac{3}{10}

x=2

Question 7

Mixing: solve $\frac{x}{20}+\frac{1}{4}=\frac{2}{5}$ for $x$ mL.

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

\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 8

Dividing quantity: solve $\frac{x}{7}-\frac{2}{7}=\frac{3}{7}$ for $x$ .

$x=1$
$x=3$
$x=5$ (correct answer)
$x=7$

\frac{x}{7} - \frac{2}{7} = \frac{3}{7}

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 9

Recipe scaling: solve $\frac{x}{3}+\frac{1}{4}=\frac{11}{12}$ for $x$ cups.

$x=1$
$x=2$ (correct answer)
$x=\frac{3}{2}$
$x=\frac{5}{2}$

\frac{x}{3} + \frac{1}{4} = \frac{11}{12}

x=2

Question 10

Mixing: solve $\frac{x}{6}+\frac{1}{2}=\frac{5}{6}$ for $x$ mL.

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

\frac{x}{6} + \frac{1}{2} = \frac{5}{6}

x=2

Question 11

Which equation is equivalent to $0.25x + \frac{1}{3} = 0.75x - \frac{2}{3}$ when all terms are expressed with integer coefficients?

$3x + 4 = 9x - 8$ (correct answer)
$25x + 33 = 75x - 67$
$x + 4 = 3x - 8$
$75x + 100 = 225x - 200$

Explanation: Convert decimals to fractions:

\frac{1}{4}x + \frac{1}{3} = \frac{3}{4}x - \frac{2}{3}

. To clear denominators, multiply by LCD = 12:

12 \cdot \frac{1}{4}x + 12 \cdot \frac{1}{3} = 12 \cdot \frac{3}{4}x - 12 \cdot \frac{2}{3}

. This gives

3x + 4 = 9x - 8

. Choice B uses 100 as multiplier unnecessarily. Choice C incorrectly multiplies the

x

terms. Choice D uses 300 as multiplier and makes coefficient errors.

Question 12

Cost sharing: solve $\frac{x}{3}+\frac{1}{9}=\frac{4}{9}$ dollars for $x$ .

$x=\frac{1}{3}$
$x=\frac{2}{3}$
$x=1$ (correct answer)
$x=\frac{4}{3}$

\frac{x}{3} + \frac{1}{9} = \frac{4}{9}

x=1

Question 13

Dividing items: solve $\frac{x}{9}-\frac{1}{9}=\frac{2}{3}$ for $x$ .

$x=5$
$x=6$
$x=7$ (correct answer)
$x=8$

\frac{x}{9} - \frac{1}{9} = \frac{2}{3}

x=7

Question 14

Recipe: solve $\frac{x}{2}+\frac{1}{3}=\frac{5}{6}$ for $x$ cups.

$x=\frac{1}{2}$
$x=1$ (correct answer)
$x=\frac{3}{2}$
$x=2$

\frac{x}{2} + \frac{1}{3} = \frac{5}{6}

x=1

Question 15

Cost sharing: solve $\frac{x}{6}+\frac{1}{12}=\frac{5}{12}$ dollars for $x$ .

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

\frac{x}{6} + \frac{1}{12} = \frac{5}{12}

x=2

Question 16

Recipe scaling: solve $\frac{x}{3}-\frac{1}{6}=\frac{1}{2}$ for $x$ cups.

$x=\frac{3}{2}$
$x=2$ (correct answer)
$x=\frac{5}{2}$
$x=3$

\frac{x}{3} - \frac{1}{6} = \frac{1}{2}

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 17

Divide snacks: solve $\frac{x}{6}+\frac{1}{3}=\frac{5}{6}$ for $x$ .

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

\frac{x}{6} + \frac{1}{3} = \frac{5}{6}

x=3

Question 18

Mixing: solve $\frac{x}{9}+\frac{2}{3}=\frac{8}{9}$ for $x$ mL.

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

\frac{x}{9} + \frac{2}{3} = \frac{8}{9}

x=2

Question 19

Distance-speed: solve $\frac{x}{9}-\frac{1}{3}=\frac{1}{6}$ for $x$ miles.

$x=\frac{3}{2}$
$x=3$
$x=\frac{9}{2}$ (correct answer)
$x=6$

\frac{x}{9} - \frac{1}{3} = \frac{1}{6}

x=\frac{9}{2}

Question 20

Cost sharing: solve $\frac{5}{6}x-10=40$ dollars for $x$ .

$x=36$
$x=60$ (correct answer)
$x=50$
$x=72$

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.