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📝 Clear fractions by multiplying by the LCD (13 MCQs)

📖 From Digital SAT Algebra • 2. Linear Equations And Inequalities • 13 questions available

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What is Clear fractions by multiplying by the LCD?

Definition:
Clearing fractions by multiplying by the LCD involves identifying the least common denominator of all fractional terms in an equation and multiplying every term on both sides by this LCD. This eliminates all denominators, converting the equation into an equivalent integer equation.

Working:
For $\frac{1}{3}x + \frac{1}{2} = \frac{5}{6}$ , the LCD of 3, 2, and 6 is 6. Multiply each term by 6: $6 \cdot \frac{1}{3}x + 6 \cdot \frac{1}{2} = 6 \cdot \frac{5}{6}$ , which gives $2x + 3 = 5$ . Then solve: $2x = 2$ , $x = 1$ .

Example:
Solve $\frac{2}{5}y - \frac{1}{10} = \frac{3}{10}$ . LCD is 10. Multiply: $10 \cdot \frac{2}{5}y - 10 \cdot \frac{1}{10} = 10 \cdot \frac{3}{10}$ , giving $4y - 1 = 3$ , add 1: $4y = 4$ , $y = 1$ .

Reason:
This technique simplifies the equation significantly, making it easier to apply standard solving steps without dealing with fractions.

Easy

Medium

Hard

📝 All Clear fractions by multiplying by the LCD MCQs

Q1. For $\frac{2x}{5} - \frac{1}{10} = 3$ , which equation results after multiplying every term by the LCD?

4x - 1 = 30

✅

2x - 1 = 30

4x - 1 = 3

2x - 5 = 30

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The LCD of 5 and 10 is 10. Multiplying the entire equation by 10 gives $10(\frac{2x}{5}) - 10(\frac{1}{10}) = 10(3)$ , simplifying to $4x - 1 = 30$ .

Q2. A student claims that multiplying $\frac{x}{4} + \frac{x}{6} = 5$ by 12 changes the equation because 12 is larger than both denominators. Which response is mathematically correct?

A.The equation changes because only the smallest denominator may be used

B.The equation remains equivalent because every term is multiplied by the same nonzero number ✅

C.The equation changes because 12 must be divided by each denominator

D.The equation remains equivalent only if

x

is positive

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Multiplying both sides of an equation by the same nonzero number preserves equality. Since 12 is a common multiple of 4 and 6, it clears both denominators without changing the solution set.

Q3. Why is the LCD generally preferred over a larger common multiple when clearing fractions from an equation?

A.A larger multiple always creates extraneous solutions

B.The LCD makes the variable disappear automatically

C.The LCD usually produces the smallest integer coefficients and a simpler equation ✅

D.A larger multiple cannot be distributed across parentheses

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Any common multiple can clear the denominators when applied to every term. However, the least common denominator usually produces smaller coefficients, reducing arithmetic complexity and making errors less likely during later steps.

Q4. Consider $\frac{x-2}{3} + \frac{x+1}{4} = 7$ . Which transformed equation correctly clears all fractions?

4(x-2) + 3(x+1) = 28

12(x-2) + 12(x+1) = 7

3(x-2) + 4(x+1) = 84

4(x-2) + 3(x+1) = 84

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The LCD of 3 and 4 is 12. Multiplying every term by 12 gives $4(x-2) + 3(x+1) = 84$ , because $12 \div 3 = 4$ , $12 \div 4 = 3$ , and $12(7) = 84$ .

Q5. A school club spends $\frac{2}{5}$ of its budget on supplies and $\frac{1}{10}$ on transportation, leaving $90. If the original budget is $B$ , which equation results after clearing fractions?

4B + B = 900

✅

4B + B = 90

2B + B = 900

2B + 5B = 900

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The spending fractions total $\frac{2}{5} + \frac{1}{10} = \frac{1}{2}$ , so $B - \frac{2B}{5} - \frac{B}{10} = 90$ . Multiplying by the LCD 10 gives $10B - 4B - B = 900$ , or $5B = 900$ .

Q6. A student solves $\frac{x}{6} - \frac{x}{4} = 2$ by multiplying only the first term by 12, obtaining $2x - \frac{x}{4} = 2$ . What is the key error?

A.The LCD should be 24, not 12

B.The student should multiply only the right side

C.The same multiplier must be applied to every term of the equation ✅

D.The variable cannot appear in two fractions

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The LCD of 6 and 4 is 12, so the chosen multiplier is correct. The mistake is applying it to only one term. Every term on both sides must be multiplied by 12 to preserve equivalence.

Q7. A student writes $\frac{x}{3} + \frac{2}{5} = 7$ , then multiplies by 15 and obtains $5x + 2 = 7$ . Which correction identifies the error?

A.The left side should be

5x + 6

, and the right side should be 105

B.The left side should be

3x + 10

, and the right side should be 35

C.The left side should be

5x + 2

, and the right side should be 105 ✅

D.The LCD should be 8, so the equation cannot be cleared

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Multiplying by 15 gives $15(\frac{x}{3}) + 15(\frac{2}{5}) = 15(7)$ . These simplify to $5x + 6 = 105$ . The student incorrectly treated $15(2/5)$ as 2 and $15(7)$ as 7.

Q8. A graph shows two lines representing the two sides of an equation. They intersect at $x=6$ . If the equation is $\frac{x}{2} + \frac{x}{3} = 5$ , what should happen after multiplying the equation by 6?

A.The intersection moves to

x=1

because the coefficients increase

B.The intersection remains at

x=6

because multiplying both sides by 6 preserves equality ✅

C.The graph becomes horizontal because fractions are removed

D.The equation gains a second intersection because the LCD introduces another solution

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The equation simplifies to $3x + 2x = 30$ , or $5x = 30$ , giving $x = 6$ . Multiplying both sides by the LCD changes the equation's appearance but not its solution, so the intersection remains unchanged.

Q9. A technician models a repair using $\frac{3}{4}t + \frac{1}{2} = 8$ . To avoid fractions, she multiplies by 4. Which equation and solution are correct?

3t + 2 = 32, t = 10

✅

3t + 1 = 32, t = \frac{31}{3}

12t + 2 = 8, t = \frac{1}{2}

3t + 2 = 8, t = 2

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The LCD is 4. Multiplying every term by 4 gives $3t + 2 = 32$ . Subtracting 2 gives $3t = 30$ , so $t = 10$ . The transformation preserves the original equation's solution.

Q10. Two students solve $\frac{x+2}{4} - \frac{x-1}{6} = 3$ . Student A multiplies by 12, while Student B multiplies by 24. Which statement best compares their methods?

A.Only Student A can obtain a valid solution

B.Only Student B can obtain a valid solution

C.Both can obtain the same solution if they multiply every term correctly ✅

D.Neither method works because the denominators contain variables

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Both 12 and 24 are common multiples of 4 and 6. Multiplying every term by either nonzero number preserves equivalence. The LCD is preferable because it usually keeps coefficients smaller.

Q11. A quantity $x$ satisfies $\frac{x}{2} + \frac{x}{3} - \frac{x}{6} = 10$ . Which result follows most efficiently after clearing fractions?

3x + 2x - x = 10

, so

x = 5

6x + 3x - 2x = 60

, so

x = 12

✅

3x + 2x - x = 60

, so

x = 20

6x + 2x - 3x = 60

, so

x = 12

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The LCD is 6. Multiplying each term by 6 produces $3x + 2x - x = 60$ , which simplifies to $4x = 60$ . Therefore $x = 15$ , showing why careful simplification after clearing fractions is essential.

Q12. For $\frac{x}{2} + \frac{x}{3} = \frac{x+5}{6} + 4$ , a student multiplies by 6 and gets $3x + 2x = x + 5 + 4$ . What should the student recognize?

A.The transformation is correct and preserves the equation

B.The right side should be

x + 5 + 24

, because

6(4)=24

✅

C.The first term should become

2x

, not

3x

D.The LCD should be 12, because three denominators are present

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Multiplying by 6 gives $3x + 2x = x + 5 + 24$ . The student correctly clears the fractional terms but incorrectly leaves 4 unchanged. Every term, including the constant 4, must be multiplied by 6.

Q13. A puzzle uses $\frac{x}{3} + \frac{x}{4} - \frac{x}{6} = 5$ . Without solving each fraction separately, which cleared equation and solution are correct?

4x + 3x - 2x = 60, x = 12

✅

12x + 12x - 12x = 5, x = 5

4x + 3x - 2x = 30, x = 6

2x + 3x - 4x = 60, x = 60

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The LCD of 3, 4, and 6 is 12. Multiplying by 12 gives $4x + 3x - 2x = 60$ . Combining terms yields $5x = 60$ , so $x = 12$ . This method avoids separately manipulating each fraction.

🔗 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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