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📝 Equations with fraction coefficients (12 MCQs)

📖 From Digital SAT Algebra • 2. Linear Equations And Inequalities • 12 questions available

←Previous Topic:Equations with Fractions or Decimals📝 Equations with fraction coefficientsClear fractions by multiplying by the LCD:Next Topic→

What is Equations with fraction coefficients?

Definition:
Equations with fraction coefficients have variables multiplied by fractions, such as $\frac{2}{3}x = 8$ or $\frac{1}{4}y + 2 = 5$ . These are solved by multiplying both sides by the reciprocal of the fraction coefficient or by clearing all fractions using the LCD.

Working:
For $\frac{3}{4}x = 9$ , multiply both sides by the reciprocal $\frac{4}{3}$ : $x = 9 \cdot \frac{4}{3} = 12$ . Alternatively, multiply by the LCD of all denominators if there are multiple terms.

Example:
Solve $\frac{2}{5}m = 14$ . Multiply both sides by $\frac{5}{2}$ : $m = 14 \cdot \frac{5}{2} = 35$ . Check: $\frac{2}{5}(35) = 14$ .

Reason:
This method directly isolates the variable when it has a fractional multiplier, avoiding complex fraction operations and providing a clean solution.

Easy

Medium

Hard

📝 All Equations with fraction coefficients MCQs

Q1. Which value of $x$ satisfies $\frac{3}{4}x - \frac{1}{2} = 7$ ?

A.8

B.9

C.10 ✅

D.11

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Adding $\frac{1}{2}$ to both sides gives $\frac{3}{4}x = \frac{15}{2}$ . Multiplying by $\frac{4}{3}$ produces $x = 10$ . The key idea is to preserve equality while eliminating the fractional coefficient.

Q2. An equation is written as $\frac{2}{3}x + 5 = 17$ . Which first step most efficiently removes the fraction without changing the solution?

A.Multiply only

x

by 3

B.Subtract 5 from both sides

C.Multiply every term by 3 ✅

D.Divide both sides by

\frac{2}{3}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Multiplying every term by 3 clears the denominator while preserving equality: $3(\frac{2}{3}x+5) = 3(17)$ . This gives $2x+15 = 51$ , making the remaining solution process simpler.

Q3. A student solves $\frac{5}{6}x - 4 = 11$ by multiplying only the fraction term by 6, obtaining $5x - 4 = 11$ . Why is this reasoning invalid?

A.The coefficient should be divided by 6

B.The constant term must also be multiplied by 6 ✅

C.The variable cannot have a fractional coefficient

D.The right side should be divided by 6 instead

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: When multiplying an equation by 6, every term on both sides must be multiplied by 6. The correct transformation is $5x - 24 = 66$ . Multiplying only one term changes the equation and therefore can change its solution.

Q4. A recipe uses $\frac{3}{4}$ cup of sugar per batch. A baker has 6 cups and uses 2 cups for another recipe. If $b$ represents the number of complete batches possible, which equation correctly models the situation?

3b - 2 = 6

\frac{3}{4}b + 2 = 6

\frac{3}{4}b = 8

\frac{3}{4}b = 6 - 2

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: After reserving 2 cups, $6-2=4$ cups remain. Each batch requires $\frac{3}{4}$ cup, so the model is $\frac{3}{4}b = 4$ . Solving gives $b = \frac{16}{3}$ , meaning 5 complete batches can actually be made.

Q5. A student claims that $\frac{1}{2}(x-6) = 10$ has solution $x = 14$ . Which reasoning identifies the error most precisely?

A.The student should subtract 6 after multiplying

B.The student should divide 10 by

\frac{1}{2}

, giving

x-6=5

C.The factor

\frac{1}{2}

must be distributed first, giving

\frac{x}{2}-3=10

✅

D.The student forgot that negative numbers cannot occur

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Distributing gives $\frac{x}{2}-3 = 10$ , so $\frac{x}{2} = 13$ and $x = 26$ . Alternatively, multiplying both sides by 2 gives $x-6 = 20$ , again producing $x=26$ .

Q6. A taxi company charges a fixed fee of $' in math mode at position 8: 3 plus \̲(̲ \frac{5}{2}$ …" style="color:#cc0000">3 plus $\frac{5}{2}$ dollars per kilometer. A passenger pays18. Which equation can be used to find the distance $d$ , and what is $d$ ?

3 + \frac{5}{2}d = 18; d = 6

✅

3 + \frac{5}{2}d = 18; d = 7.5

5 + \frac{3}{2}d = 18; d = 8

\frac{5}{2}(d+3) = 18; d = 4.2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The fixed fee is $3 and the distance charge is $\frac{5}{2}d$ , so $3 + \frac{5}{2}d = 18$ . Subtracting 3 gives $\frac{5}{2}d = 15$ , and multiplying by $\frac{2}{5}$ gives $d = 6$ kilometers.

Q7. Consider the equation $\frac{4}{5}x + 2 = 10$ . Which sequence of operations correctly isolates $x$ ?

A.Subtract 2, then multiply by

\frac{4}{5}

B.Subtract 2, then multiply by

\frac{5}{4}

✅

C.Multiply by

\frac{4}{5}

, then subtract 2

D.Divide by 5, then multiply by 4

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: First subtract 2 from both sides to obtain $\frac{4}{5}x = 8$ . Since $x$ is multiplied by $\frac{4}{5}$ , multiply both sides by its reciprocal $\frac{5}{4}$ . Therefore $x = 10$ .

Q8. A graph of $y = \frac{2}{3}x + 4$ and the horizontal line $y = 10$ intersect at one point. What $x$ -coordinate should the intersection have?

A.6

B.8

C.9 ✅

D.12

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: At the intersection, the two $y$ -values are equal, so $\frac{2}{3}x + 4 = 10$ . Subtracting 4 gives $\frac{2}{3}x = 6$ . Multiplying by $\frac{3}{2}$ gives $x = 9$ , so the intersection has $x$ -coordinate 9.

Q9. Two students solve $\frac{3}{4}x + 2 = \frac{1}{2}x + 8$ . Student A multiplies the entire equation by 4 first. Student B subtracts $\frac{1}{2}x$ first. Which statement is correct?

A.Only Student A can obtain the correct solution

B.Only Student B can obtain the correct solution

C.Both methods are valid if every operation is applied to both sides ✅

D.Neither method is valid because fractions cannot be removed

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Both approaches preserve equality. Student A obtains $3x+8=2x+32$ , while Student B obtains $\frac{1}{4}x+2=8$ . Both lead to $x=24$ , demonstrating that different valid sequences can produce the same solution.

Q10. A student transforms $\frac{7}{8}x - 3 = \frac{1}{4}x + 9$ into $7x - 3 = 2x + 9$ by multiplying by 8. What should the corrected equation be, and what solution follows?

7x - 24 = 2x + 72, x = 19.2

✅

7x - 3 = 2x + 9, x = 2

7x - 24 = 2x + 9, x = 6.6

7x - 24 = 2x + 72, x = 16

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Multiplying the original equation by 8 requires multiplying every term, including the constants. Thus $7x - 24 = 2x + 72$ . Subtracting $2x$ gives $5x = 96$ , so $x = \frac{96}{5} = 19.2$ .

Q11. A number is increased by $\frac{2}{3}$ of itself, and the result is 40. Another student argues that the number must be 24 because $\frac{2}{3} \times 40 = 24$ . What is the actual number?

A.15

B.20

C.24 ✅

D.30

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Let the number be $x$ . The statement means $x + \frac{2}{3}x = 40$ , not $\frac{2}{3}(40) = x$ . Combining terms gives $\frac{5}{3}x = 40$ , and multiplying by $\frac{3}{5}$ yields $x = 24$ . Therefore option C is actually the correct value, while the distractor reasoning happens to reach it by coincidence.

Q12. What is the most useful first step for solving $\frac{x}{3} + \frac{5}{6} = 4$ by clearing fractions?

A.Multiply every term by 6 ✅

B.Multiply only the fractions by 6

C.Multiply every term by 3

D.Add the denominators 3 and 6

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The denominators are 3 and 6, so their least common denominator is 6. Multiplying every term by 6 removes both fractions while preserving equality, producing a simpler linear equation.

🔗 Related Topics (MCQs)

📝 Subtraction and Addition Properties of Equality MCQs 📝 How to verify a solution to an equation MCQs 📝 Subtraction Property of Equality MCQs 📝 Addition Property of Equality MCQs 📝 Solving equations with simplification MCQs

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