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📝 How to verify a solution to an equation (14 MCQs)

📖 From Digital SAT Algebra • 2. Linear Equations And Inequalities • 14 questions available

What is How to verify a solution to an equation?

Definition:
Verifying a solution to an equation means substituting the proposed value of the variable back into the original equation to check if it makes the equation a true statement. If the left-hand side equals the right-hand side after substitution, the value is a correct solution; otherwise, it is not.

Working:
To verify, replace the variable with the number and simplify both sides independently using arithmetic operations. If the simplified values are equal, the solution is confirmed. For example, to verify x=4x = 4 for 2x+1=92x + 1 = 9, substitute x=4x = 4: 2(4)+1=8+1=92(4) + 1 = 8 + 1 = 9, which matches the right-hand side.

Example:
Verify if t=2t = -2 is a solution to 3t4=103t - 4 = -10. Substitute: 3(2)4=64=103(-2) - 4 = -6 - 4 = -10. Both sides equal 10-10, so t=2t = -2 is verified as a solution.

Reason:
Verification is essential to catch arithmetic errors and ensure the answer actually satisfies the original equation. It provides a reliable check that boosts confidence in the solution process.

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Easy
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Medium
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Hard

📝 All How to verify a solution to an equation MCQs

Q1. A student claims that x=3x = -3 is a solution to 2(x+5)3(x2)=7x2(x+5) - 3(x-2) = 7 - x. Without fully simplifying, which single substitution check would most efficiently disprove the claim?

A.Substitute x=3x=-3 into 2(x+5)2(x+5) only.
B.Substitute x=3x=-3 into 7x7-x only.
C.Substitute x=3x=-3 into the left side and right side separately and compare. ✅
D.Substitute x=3x=-3 into x+5x+5 and x2x-2 only.
💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Efficient verification requires comparing the simplified numerical values of both sides after substitution. Option C is the only one that checks the entire equation. Substituting into only parts (A, B, D) is incomplete and cannot disprove the equality because errors in other terms might cancel or be missed, leading to a false conclusion.

Q2. Given the equation 4(2y1)=3(y+5)2y4(2y-1) = 3(y+5) - 2y. If you simplify the right side to y+15y+15, which of the following is the best next step to verify if y=2y=2 is a solution?

A.Substitute y=2y=2 into the original left and simplified right side.
B.Substitute y=2y=2 into the original equation only.
C.Substitute y=2y=2 into the simplified equation 4(2y1)=y+154(2y-1)=y+15.
D.Both B and C are equally valid and sufficient. ✅
💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Since the simplified equation 4(2y1)=y+154(2y-1)=y+15 is mathematically equivalent to the original, substituting into either the original (B) or the simplified form (C) is valid. Option A is also correct but redundant. Option D correctly identifies that both B and C are acceptable verification methods, demonstrating understanding that simplification preserves equality.

Q3. Maria solved an equation and got x=4x=4. To verify, she substituted x=4x=4 into the left side and got 12, and into the right side and got 12. She concluded the solution is correct. Is her conclusion always valid, and why?

A.Yes, because equal numerical values on both sides confirm the solution.
B.No, because she might have made an arithmetic error in substitution.
C.Yes, because verification only requires checking the final value.
D.No, because she must also verify that no division by zero occurred in the original equation. ✅
💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: While equal values confirm the equality at x=4x=4, verification must also ensure that the solution does not make any denominator zero or violate the domain of the original equation. Maria’s check is necessary but not sufficient for equations with variables in denominators. Option D highlights this critical, often-overlooked step in verification, which is a higher-order error analysis.

Q4. The graph of y=3x5y = 3x - 5 and y=2x+10y = -2x + 10 intersect at the point (3,4)(3, 4). Which of the following statements is true about the equation 3x5=2x+103x - 5 = -2x + 10?

A.x=3x=3 is a solution because the y-values are equal at x=3x=3. ✅
B.x=4x=4 is a solution because the y-value is 4.
C.The solution is x=4x=4 because the intersection point has y-coordinate 4.
D.The equation has no solution because the lines intersect.
💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The intersection point (3,4)(3,4) means that when x=3x=3, both expressions yield y=4y=4. Therefore, x=3x=3 satisfies the equation 3x5=2x+103x-5 = -2x+10. Option A correctly interprets the graph. Options B and C confuse the x and y coordinates of the intersection point, a common misconception. Option D is false because intersection implies a solution exists.

Q5. A teacher writes 5x3=2x+95x - 3 = 2x + 9 on the board. Four students verify x=4x=4 as follows. Who used the most rigorous verification method?

A.Alice: 'I solved it and got 4, so it's correct.'
B.Bob: 'I plugged in 4: LHS=17, RHS=17, so it works.' ✅
C.Carol: 'I checked that both sides are equal when x=4x=4.'
D.David: 'I graphed both sides and they intersect at (4,17).'
💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Bob’s method is the most rigorous because he explicitly computes both sides numerically and compares them. Carol’s statement is vague without computation. Alice relies on her solving process, which may have errors. David’s graph is a visual approximation, not an exact algebraic verification unless exact coordinates are given. Bob provides concrete, verifiable numerical evidence, making his method the strongest.

Q6. Consider the equation x+2x1=3\frac{x+2}{x-1} = 3. A student verifies x=2.5x = 2.5 by substituting: 2.5+22.51=4.51.5=3\frac{2.5+2}{2.5-1} = \frac{4.5}{1.5} = 3. They conclude the solution is correct. Which critical step did they omit?

A.Checking if the left side simplifies to 3.
B.Checking if the denominator is zero for any other value.
C.Checking if x=2.5x=2.5 makes the denominator zero. ✅
D.Checking if the right side is defined for x=2.5x=2.5.
💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: While the substitution works, the student must verify that the denominator x1x-1 is not zero at x=2.5x=2.5. Here, 2.51=1.502.5-1=1.5 \neq 0, so it's safe. However, the omission of this check is a critical error in verification, especially for rational equations. Option C correctly identifies this missing step, emphasizing that domain restrictions are part of solution verification, not just numerical equality.

Q7. Two equations are given: (I) 2x+3=72x + 3 = 7 and (II) 4x+6=144x + 6 = 14. If x=2x=2 is verified as a solution for (I), what can you conclude about (II) without solving it?

A.x=2x=2 is also a solution because (II) is (I) multiplied by 2. ✅
B.x=2x=2 is not necessarily a solution because the equations are different.
C.x=2x=2 is a solution only if we verify it separately.
D.x=2x=2 cannot be a solution for (II) because the coefficients changed.
💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Equation (II) is exactly 2 times Equation (I). Since multiplying an equation by a non-zero constant produces an equivalent equation, any solution of (I) is also a solution of (II). Option A demonstrates this equivalence property. Options B, C, and D reflect misconceptions about equivalent equations, failing to recognize the scalar multiple relationship, which is a key algebraic concept.

Q8. A student claims that x=2x = -2 is a solution to x+1=3|x+1| = 3. Their verification: 2+1=1=1|-2+1| = |-1| = 1, which is not 3. They conclude x=2x=-2 is not a solution. Is their reasoning and conclusion correct?

A.Yes, because substitution disproved it. ✅
B.No, because absolute value equations have two solutions.
C.Yes, but they should also check x=2x=2.
D.No, because they evaluated the absolute value incorrectly.
💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The student correctly substituted x=2x=-2 into the absolute value expression and obtained 1, which does not equal 3. Therefore, x=2x=-2 is indeed not a solution. Option A is correct. Options B and C are distractors: while x+1=3|x+1|=3 has solutions x=2x=2 and x=4x=-4, that doesn't make x=2x=-2 a solution. Option D is false because the evaluation 1=1|-1|=1 is correct. This tests whether students understand that verification is a definitive test.

Q9. Which of the following is NOT a valid way to verify if x=5x=5 is a solution to 3(x2)+4=2x+73(x-2) + 4 = 2x + 7?

A.Substitute x=5x=5 into the original equation and check if both sides equal.
B.Simplify the equation to 3x2=2x+73x-2 = 2x+7, then substitute x=5x=5.
C.Substitute x=5x=5 into the left side only and see if it equals the right side's value at x=5x=5.
D.Substitute x=5x=5 into 3(x2)3(x-2) and 2x+72x+7, ignoring the constants +4. ✅
💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Options A, B, and C are all valid verification methods because they maintain the equality or compare equivalent forms. Option D is invalid because it ignores the constant term +4 on the left side, changing the equation. Substituting into only parts of an expression is a common error. This question requires students to analyze the validity of different verification procedures and identify an incomplete or incorrect method, testing error analysis skills.

Q10. The equation ax+b=cx+dax + b = cx + d has a solution x=3x = 3. If you multiply both sides by 0, you get 0=00=0. Is x=3x=3 still a solution to the new equation 0=00=0?

A.Yes, because multiplying by 0 preserves all solutions.
B.No, because multiplying by 0 is not allowed in algebra.
C.Yes, but the new equation has infinitely many solutions, so x=3x=3 is one of them.
D.No, because the new equation is not equivalent to the original. ✅
💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Multiplying an equation by 0 yields the identity 0=00=0, which is true for all xx, including x=3x=3. However, this operation is not reversible (it loses information), so the new equation is not equivalent to the original. While x=3x=3 satisfies 0=00=0, it cannot be verified as a solution to the original equation using this transformed equation. Option D correctly distinguishes between 'satisfying the new equation' and 'being a solution to the original', a subtle and advanced concept.

Q11. A student simplifies 2(x+3)=62(x+3) = 6 to 2x+6=62x+6=6, then to 2x=02x=0, and finally x=0x=0. They verify by substituting x=0x=0 into the original: LHS=6, RHS=6. Their verification is valid, but what is the deeper purpose of this step?

A.To confirm the arithmetic in the solving steps.
B.To check that no extraneous solutions were introduced during simplification.
C.To ensure the solution satisfies the original equation, not just the simplified one.
D.Both A and C are correct. ✅
💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Verification serves two purposes: (1) it checks for arithmetic errors made during solving (Option A), and (2) it ensures the final value satisfies the original equation, especially when operations like multiplying by a variable or squaring might introduce extraneous solutions (Option C). Option D correctly combines both, showing a comprehensive understanding of why verification is essential, beyond just a routine check.

Q12. Given the system y=2x+1y = 2x+1 and y=x+4y = -x+4, a student finds the intersection at (1,3). To verify the solution for the equation 2x+1=x+42x+1 = -x+4, they substitute x=1x=1 into both linear expressions. Which of the following correctly interprets the verification?

A.Substitution confirms that x=1x=1 makes the expressions equal, so it's a solution.
B.Substitution confirms that the point (1,3) lies on both lines, so x=1x=1 is a solution.
C.Both A and B are correct interpretations. ✅
D.Substitution only verifies the first equation, not the equality.
💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The equation 2x+1=x+42x+1 = -x+4 represents the x-coordinate of the intersection of the two lines. Substituting x=1x=1 gives 3=33=3, verifying the equality (Option A). Graphically, this means the point (1,3) lies on both lines, confirming the intersection (Option B). Option C correctly combines both interpretations. This question requires students to connect algebraic verification with graphical meaning, testing mixed concepts and graph interpretation.

Q13. A student verifies x=2x=2 for 3x2=43x - 2 = 4 by computing LHS = 3(2)2=43(2)-2 = 4 and RHS = 4. They then state, 'Since LHS = RHS, the solution is verified.' Another student says, 'But you assumed the solution to check it, that's circular reasoning.' Who is correct?

A.The first student, because verification is exactly that check.
B.The second student, because you can't use the solution to prove itself.
C.Both are partially correct: verification is not proof, but it's a necessary check. ✅
D.Neither, because verification is never needed if you solved correctly.
💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Verification is not a proof of the solution's uniqueness or derivation; it's a confirmation that the proposed value satisfies the equation. The first student is correct that the check works. The second student raises a valid philosophical point: verification doesn't prove the solving steps were correct, but it does provide evidence. Option C captures this nuance: verification is a necessary (but not sufficient) check for correctness, not a circular argument. This tests deep understanding of the role of verification in mathematics.

Q14. A student solved 2x+3=x\sqrt{2x+3} = x and got solutions x=3x=3 and x=1x=-1. They verified x=3x=3: 9=3\sqrt{9}=3, works. For x=1x=-1: 1=1\sqrt{1} = -1, which is false, so they reject x=1x=-1. Their verification process is correct, but what is the fundamental reason x=1x=-1 failed?

A.Arithmetic error in evaluating the square root.
B.Squaring both sides introduced an extraneous solution.
C.The equation has no negative solutions because of the square root.
D.Both B and C are correct. ✅
💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: Squaring both sides of an equation can introduce extraneous solutions that satisfy the squared equation but not the original. Here, x=1x=-1 satisfies (2x+3)2=x2(\sqrt{2x+3})^2 = x^2 but not the original because the principal square root is non-negative, so 1=11\sqrt{1}=1 \neq -1. Option B is the algebraic reason, and Option C is the domain/logical reason. Option D correctly combines both, requiring students to understand the underlying cause of the failed verification, not just the procedural check.

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