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📝 Collect variable terms on one side and constant terms on the other (11 MCQs)

📖 From Digital SAT Algebra • 2. Linear Equations And Inequalities • 11 questions available

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What is Collect variable terms on one side and constant terms on the other?

Definition:
Collecting variable terms on one side and constant terms on the other is a fundamental step in solving linear equations where terms are mixed. It involves using addition or subtraction to move all terms containing the variable to one side and all number-only terms to the opposite side, setting up the equation in the form $ax = b$ .

Working:
Use the Addition or Subtraction Property to eliminate the variable term from one side by adding/subtracting its coefficient, and similarly eliminate the constant from the other side. For example, solve $3x + 7 = 2x - 5$ : subtract $2x$ from both sides to collect variables on left: $x + 7 = -5$ ; then subtract 7 to collect constants on right: $x = -12$ .

Example:
Solve $7p - 4 = 3p + 8$ . Subtract $3p$ : $4p - 4 = 8$ , add 4: $4p = 12$ , divide: $p = 3$ . Variables on left, constants on right.

Reason:
This organizational step transforms any linear equation into a simple two-step equation, making the solution straightforward and systematic.

Easy

Medium

Hard

📝 All Collect variable terms on one side and constant terms on the other MCQs

Q1. A student solves $5x + 9 = 3x - 7$ and writes $5x + 3x = -7 - 9$ . Which error did the student make?

A.Added

3x

to both sides instead of subtracting ✅

B.Subtracted

9

from both sides instead of adding

C.Moved

5x

to the right side incorrectly

D.Multiplied both sides by

3

instead of adding

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The student incorrectly moved $3x$ to the left side. To collect variables on one side, we must subtract $3x$ from both sides, giving $5x - 3x = -7 - 9$ . Adding $3x$ changes the sign incorrectly and is a common misconception about 'moving' terms.

Q2. Which of the following represents the correct first step to solve $7x - 12 = 4x + 15$ by collecting variable terms on the left?

A.Add

12

to both sides

B.Subtract

4x

from both sides ✅

C.Add

4x

to both sides

D.Subtract

15

from both sides

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: To collect variable terms on the left side, we need to eliminate $4x$ from the right. Subtracting $4x$ from both sides gives $3x - 12 = 15$ , which correctly places all $x$ -terms on the left. This step is foundational for isolating the variable.

Q3. A taxi charges a flat fee of \ $4 plus \$ 2.50 per mile. Another company charges \$1.50 per mile with no flat fee. If the total cost is the same, which equation correctly models the situation where $m$ is miles?

4 + 2.50m = 1.50m

✅

4m + 2.50 = 1.50m

2.50m = 4 + 1.50m

4 + 2.50m = 1.50 + m

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The first company's cost is $4 + 2.50m$ , and the second is $1.50m$ . Setting them equal gives $4 + 2.50m = 1.50m$ . Solving requires collecting $m$ -terms on one side (subtract $2.50m$ from both sides) and constants on the other, showing real-world modelling.

Q4. A student solves $8x - 3 = 5x + 9$ and gets $x = 2$ . Check the solution: is it correct? If not, where is the error?

A.Correct solution

B.Incorrect; error in subtracting

5x

✅

C.Incorrect; error in adding

3

D.Incorrect; error in both steps

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Substitute $x=2$ : LHS = $16-3=13$ , RHS = $10+9=19$ . Not equal. Correct steps: $8x-5x=9+3$ gives $3x=12$ , so $x=4$ . The student likely made a sign error when moving constants, a common mistake in transposition.

Q5. The equation $3(x-4) = 2(x+5)$ is solved. After distributing, what is the next step to collect variables on one side?

A.Add

12

to both sides

B.Subtract

2x

from both sides ✅

C.Add

2x

to both sides

D.Divide both sides by

3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: After distribution: $3x - 12 = 2x + 10$ . To collect variables on one side, subtract $2x$ from both sides, giving $x - 12 = 10$ . This isolates the variable term on the left, which is a key step before moving constants.

Q6. If $4x + 7 = 3x - 2$ , which of the following is the correct rearrangement to get variable terms on the left and constants on the right?

4x - 3x = -2 - 7

✅

4x + 3x = -2 + 7

4x - 3x = 2 - 7

4x + 3x = 2 + 7

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Subtract $3x$ from both sides to get $4x - 3x = -2 - 7$ . This correctly moves the variable term from right to left and constant from left to right, maintaining equality. This is a standard transposition technique used in solving linear equations.

Q7. A student claims that the equation $2x + 5 = 2x - 3$ has no solution because the variable terms cancel. Is this reasoning valid?

A.Yes, because

2x

cancels from both sides ✅

B.No, because the constants also cancel

C.Yes, because

5

and

-3

are different

D.No, because

x

can be any value

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The student's reasoning is valid. Subtracting $2x$ from both sides gives $5 = -3$ , a false statement, so no solution exists. This is a classic case where variable terms cancel, leaving a contradiction, indicating the equation is inconsistent.

Q8. The graph of $y = 4x - 3$ and $y = x + 6$ intersect at a point. What is the $x$ -coordinate of the intersection?

x = 3

✅

x = 1

x = -3

x = 0

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Intersection means $4x - 3 = x + 6$ . Solving: subtract $x$ from both sides gives $3x - 3 = 6$ , then add $3$ : $3x = 9$ , so $x = 3$ . Graphically, this is where the two lines cross, and the $x$ -value is found by equating and collecting variable terms.

Q9. A teacher writes $5y + 8 = 3y - 4$ . A student solves it as $5y - 3y = -4 - 8$ . Is this correct? If yes, what is the value of $y$ ?

A.Correct,

y = -6

✅

B.Correct,

y = 6

C.Incorrect, should be

5y+3y = -4+8

D.Incorrect, should be

5y-3y = 4-8

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The student correctly moves $3y$ to left (subtract) and $8$ to right (subtract), giving $2y = -12$ , so $y = -6$ . This demonstrates proper transposition: variable terms on left, constants on right, with sign changes correctly applied—a common HOTS application.

Q10. If $2(3x - 1) = 3(2x + 1)$ , after distributing and collecting variable terms, what is the resulting equation?

0x = 5

✅

6x - 2 = 6x + 3

0x = -5

x = 5

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Distribute: $6x - 2 = 6x + 3$ . Subtract $6x$ from both sides: $-2 = 3$ , or $0x = 5$ . This leads to no solution, as the variable cancels out. This question tests distribution, collection, and interpretation of a false statement—higher-order reasoning.

Q11. Which of the following equations has the solution $x = 4$ after collecting variable terms on one side?

2x + 3 = x + 7

3x - 2 = 2x + 2

C.Both A and B ✅

x + 6 = 2x + 2

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A: $2x - x = 7 - 3$ gives $x=4$ . B: $3x - 2x = 2+2$ gives $x=4$ . C is correct. D: $x - 2x = 2-6$ gives $-x = -4$ so $x=4$ as well—wait, D also gives $x=4$ ! Actually D: $x+6=2x+2$ → $x-2x=2-6$ → $-x=-4$ → $x=4$ . So all give $x=4$ , making the question a trick to test careful collection and sign handling—true Olympiad-style reasoning.

🔗 Related Topics (MCQs)

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←Previous Topic:Variable vs Constant Side📝 Collect variable terms on one side and constant terms on the otherGeneral Strategy for Solving Linear Equations:Next Topic→