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📝 Solve equations with variables and constants on both sides (13 MCQs)

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

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What is Solve equations with variables and constants on both sides?

Definition:
Solving equations with variables and constants on both sides is a comprehensive method where both variable terms and constant terms appear on both sides. The strategy is to move variable terms to one side and constants to the other, then isolate the variable using division or multiplication.

Working:
First, decide which side will hold the variables (usually the side with the larger coefficient). Subtract or add variable terms to collect them on that side, and similarly move constants to the opposite side. Simplify and divide by the coefficient. For example, solve $4x - 6 = 2x + 8$ : subtract $2x$ : $2x - 6 = 8$ , add 6: $2x = 14$ , divide by 2: $x = 7$ .

Example:
Solve $6m + 1 = 4m - 9$ . Subtract $4m$ : $2m + 1 = -9$ , subtract 1: $2m = -10$ , divide by 2: $m = -5$ . Check: $6(-5)+1 = -30+1 = -29$ , $4(-5)-9 = -20-9 = -29$ .

Reason:
This general strategy handles any linear equation by systematically eliminating terms, ensuring a clear path to the solution regardless of term placement.

Easy

Medium

Hard

📝 All Solve equations with variables and constants on both sides MCQs

Q1. A student solves $5x - 8 = 3x + 12$ and writes $x = 10$ . If the student checks by substituting back, which statement is true?

A.The check gives

42 = 42

, so the solution is correct ✅

B.The check gives

50 = 50

, so the solution is correct

C.The check gives

42 = 42

, but the solution is wrong due to a sign error

D.The check gives

50 = 42

, so the solution is incorrect

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Substituting $x = 10$ into the original equation gives $5(10)-8 = 50-8=42$ and $3(10)+12=30+12=42$ . Since both sides equal 42, the solution satisfies the equation perfectly. This confirms that the student's answer is correct, and the check is the ultimate validation of solving.

Q2. Given the equation $7x + 5 = 4x - 7$ , a student subtracts $4x$ from both sides, then subtracts 5 from both sides, obtaining $3x = -12$ , so $x = -4$ . However, another student adds $7$ to both sides first. Will the final answer differ, and why?

A.Yes, the final answer will be different because the order of operations changes the solution.

B.No, the final answer will be the same because inverse operations are applied correctly in any order. ✅

C.Yes, the final answer will be different because adding 7 first changes the variable coefficient.

D.No, the final answer will be the same but only if you multiply by -1 at the end.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The order of applying inverse operations (addition/subtraction and multiplication/division) does not affect the final solution, as long as each step maintains equality. Both methods are valid sequences of inverse operations that isolate the variable. The equation is linear, so the solution set is unique and invariant under correct algebraic manipulation.

Q3. A rectangle's length is $3x + 4$ cm and its width is $2x - 1$ cm. If the perimeter is 46 cm, which equation correctly models this situation?

2(3x+4) + 2(2x-1) = 46

✅

3x+4 + 2x-1 = 46

2(3x+4)(2x-1) = 46

3x+4 = 2(2x-1) + 46

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The perimeter of a rectangle is given by $P = 2L + 2W$ . Substituting $L = 3x+4$ and $W = 2x-1$ gives $2(3x+4) + 2(2x-1) = 46$ . This equation correctly sets up the relationship. Simplifying leads to $10x+6=46$ , so $x=4$ , making length 16 cm and width 7 cm, which checks out.

Q4. A student solves $4(2x - 3) = 6x + 10$ as follows: Step 1: $8x - 12 = 6x + 10$ . Step 2: $8x - 6x = 10 + 12$ . Step 3: $2x = 22$ . Step 4: $x = 11$ . If the student's answer is incorrect, identify the first error.

A.No error; the solution

x = 11

is correct.

B.Error in Step 1: distributing 4 incorrectly.

C.Error in Step 2: moving terms changes signs incorrectly.

D.Error in Step 3: combining like terms gives

2x = -2

, not 22. ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Let's check the student's work: Step 1 is correct ( $8x-12=6x+10$ ). In Step 2, moving $6x$ to left gives $8x-6x = 10 + 12$ is actually correct if we consider adding 12 to both sides: $8x-12+12=6x+10+12$ gives $8x=6x+22$ , then subtract $6x$ gives $2x=22$ . So the error is not in sign; the real error is in Step 3: $8x-6x = 2x$ , and $10+12 = 22$ , so $2x=22$ is correct. Actually, the student's solution $x=11$ is correct. Substituting: $4(22-3)=4(19)=76$ , and $6(11)+10=66+10=76$ . So no error; the answer is correct.

Q5. The graph of $y = 2x + 5$ and $y = -3x + 20$ intersect at a point. What is the x-coordinate of the intersection?

x = 3

✅

x = 5

x = -3

x = 15

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The intersection point satisfies both equations, so set $2x+5 = -3x+20$ . Solving: add $3x$ to both sides gives $5x+5=20$ , subtract 5 gives $5x=15$ , so $x=3$ . Substituting back gives $y=11$ . The graph would show two lines crossing at (3,11), so the x-coordinate is 3. This connects algebraic solution to graphical interpretation.

Q6. Two equations are given: $3x - 7 = 5x + 1$ and $2(x - 4) = 6 - x$ . Which statement about their solutions is true?

A.Both equations have the same solution,

x = -4

. ✅

B.Both equations have the same solution,

x = 4

C.The first has solution

x = -4

, the second has

x = \frac{14}{3}

D.The first has solution

x = 4

, the second has

x = -4

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Solve the first: $3x-7=5x+1$ → subtract $3x$ : $-7=2x+1$ → subtract 1: $-8=2x$ → $x=-4$ . Solve the second: $2x-8=6-x$ → add $x$ : $3x-8=6$ → add 8: $3x=14$ → $x = 14/3$ . Wait, that is not -4. Let's re-evaluate: second equation $2(x-4)=6-x$ → $2x-8=6-x$ → add x: $3x-8=6$ → add 8: $3x=14$ → $x=14/3$ . So first gives -4, second gives 14/3. The correct option is C. However, I must check if I miscomputed first: $3x-7=5x+1$ → subtract 3x: -7 = 2x+1 → subtract 1: -8=2x → x=-4. Yes. So C is correct.

Q7. A mobile phone plan charges a flat fee of $20 plus$ 0.10 per text. Another plan charges $15 plus$ 0.15 per text. For how many texts will the two plans cost the same?

A.100 texts ✅

B.200 texts

C.50 texts

D.150 texts

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Let $x$ be the number of texts. Plan A cost: $20 + 0.10x$ . Plan B cost: $15 + 0.15x$ . Set equal: $20 + 0.10x = 15 + 0.15x$ . Subtract $0.10x$ : $20 = 15 + 0.05x$ . Subtract 15: $5 = 0.05x$ . Divide by 0.05: $x = 100$ . So at 100 texts, both plans cost $30. This is a real-world application of solving equations with variables on both sides.

Q8. A student claims that the equation $2x + 5 = 2x + 10$ has no solution because the variable terms cancel, leaving $5=10$ , which is false. Is the student correct?

A.Yes, because the equation simplifies to a false statement, so no solution. ✅

B.No, because the equation simplifies to

5=10

, which is true, so infinite solutions.

C.Yes, because any value of x makes the equation true.

D.No, because you can add

-2x

to both sides to get

5=10

, so x=0 is a solution.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Subtracting $2x$ from both sides gives $5 = 10$ , a contradiction. Since no value of $x$ can make $5$ equal to $10$ , the equation has no solution. This is a classic case of inconsistent equations where the variable cancels out and leaves a false statement. The student's reasoning is correct.

Q9. What is the solution set for the equation $3(x - 2) + 4 = 3x - 2$ ?

x = 0

x = 2

C.No solution

D.All real numbers ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Expand: $3x - 6 + 4 = 3x - 2$ → $3x - 2 = 3x - 2$ . This is an identity; it holds for every real number $x$ . The equation simplifies to the same expression on both sides, so the solution set is all real numbers. This is a higher-order problem because students must recognize that the equation is an identity rather than a conditional equation, and not jump to a numerical answer.

Q10. Given the equation $\frac{2x - 1}{3} = \frac{x + 4}{2}$ , a student multiplies both sides by 6 to get $2(2x-1) = 3(x+4)$ . Then they solve to get $x=14$ . However, when they graph $y = \frac{2x-1}{3}$ and $y = \frac{x+4}{2}$ , they see the lines intersect at $x=14$ . Is the student's algebraic solution consistent with the graph?

A.Yes, because the intersection point confirms the algebraic solution. ✅

B.No, because the graph should intersect at

x = -14

C.No, because the graph shows parallel lines, so no intersection.

D.Yes, but the graph also shows another intersection due to non-linearity.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: After multiplying by 6, we get $4x - 2 = 3x + 12$ , so $x = 14$ . Substituting $x=14$ into both sides: LHS = $(28-1)/3 = 27/3 = 9$ , RHS = $(14+4)/2 = 18/2 = 9$ . So the point (14,9) is on both lines. The graph of two non-parallel lines (slopes 2/3 and 1/2) will intersect exactly once, at this point. So the graph confirms the solution.

Q11. A student solves $4x - 7 = 3x + 5$ by first adding 7 to both sides, obtaining $4x = 3x + 12$ , then subtracting $3x$ to get $x = 12$ . Another student first subtracts $3x$ , getting $x - 7 = 5$ , then adds 7 to get $x = 12$ . Which method is more efficient, and why?

A.The first method is more efficient because it eliminates the constant on the left first.

B.The second method is more efficient because it reduces the number of steps.

C.Both methods are equally valid and yield the same result; efficiency depends on preference. ✅

D.Neither method is correct because the solution should be

x = 2

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Both methods apply inverse operations in a different order but maintain equality at each step. The first method: $4x-7+7=3x+5+7$ → $4x=3x+12$ → $x=12$ . The second: $4x-7-3x=3x+5-3x$ → $x-7=5$ → $x=12$ . Both are correct and require two steps. Efficiency is subjective; some prefer to move variables first, others constants. The key is that both are valid and lead to the same unique solution.

Q12. A triangle has angles $(2x+10)^\circ$ , $(3x-20)^\circ$ , and $(5x+30)^\circ$ . Which equation correctly represents the sum of angles, and what is the value of $x$ ?

(2x+10)+(3x-20)+(5x+30)=180

x=16

✅

(2x+10)+(3x-20)+(5x+30)=180

x=20

(2x+10)+(3x-20)+(5x+30)=360

x=34

(2x+10)+(3x-20)+(5x+30)=90

x=10

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The sum of interior angles in any triangle is $180^\circ$ . So we set up: $(2x+10)+(3x-20)+(5x+30)=180$ . Combine like terms: $10x+20=180$ → $10x=160$ → $x=16$ . Then angles are 42°, 28°, and 110°, which sum to 180°. The other options incorrectly use 360° (for quadrilaterals) or 90° (for right triangles), leading to wrong x values.

Q13. An equation is given as $5(x - 3) = 2x + 3$ . A student solves it and gets $x = 6$ . Another student checks by substituting $x = 6$ and finds LHS=15, RHS=15, so confirms. But a third student claims there is a mistake because when they graph $y = 5(x-3)$ and $y = 2x+3$ , the lines intersect at $x = -6$ . Who is correct?

A.The first student is correct;

x=6

is the solution.

B.The second student is correct because substitution confirms.

C.The third student is correct; the graph shows intersection at

x = -6

D.Both first and second are correct, but the third made a graphing error. ✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Solve algebraically: $5x - 15 = 2x + 3$ → $3x = 18$ → $x=6$ . Substitute: LHS = $5(6-3)=15$ , RHS = $2(6)+3=15$ . So $x=6$ is correct. The third student likely made a sign error while graphing or solving. The lines are not parallel and intersect at exactly one point, which is (6,15). Therefore, the first and second students are correct; the third is wrong due to a graphing misconception.

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←Previous Topic:Solve equations with variables on both sides📝 Solve equations with variables and constants on both sidesVariable vs Constant Side:Next Topic→