Checking Solutions by Substitution - Math 1

A student solves $\sqrt{x + 5} = x - 1$ and gets $x = 4$ . When checking this solution, the student finds $\sqrt{4 + 5} = \sqrt{9} = 3$ and $4 - 1 = 3$ , so both sides equal $3$ . What should the student conclude?
Answer: The solution $x = 4$ is correct since both sides equal $3$ when substituted into the original equation When solving radical equations, your primary goal is to find values that satisfy the original equation. Once you've algebraically solved for a variable, the crucial step is substituting back into the original equation to verify your solution works. The student correctly verified that $x = 4$ satisfies $\sqrt{x + 5} = x - 1$ . When substituting: $\sqrt{4 + 5} = \sqrt{9} = 3$ and $4 - 1 = 3$ . Since both sides equal 3, the solution is valid and complete. Let's examine why the other options contain flawed reasoning. Choice A suggests checking $x = -1$ , but there's no mathematical basis for this specific value. You only need to check solutions that arise from your algebraic work. Choice B incorrectly claims that $\sqrt{9} = ±3$ . This is a common misconception—the radical symbol $\sqrt{9}$ represents only the principal (positive) square root, which is 3. The equation $x^2 = 9$ has solutions $x = ±3$ , but $\sqrt{9} = 3$ only. Choice C mentions domain restrictions, but while $x + 5 ≥ 0$ is correct for the radical, requiring $x - 1 ≥ 0$ is unnecessary since the right side can be negative. The verification process confirms $x = 4$ is the complete and correct solution. Study tip: When solving radical equations, always substitute your solutions back into the original equation. This catches extraneous solutions that sometimes arise from squaring both sides, and confirms your answer is mathematically sound.
When checking if $x = -2$ is a solution to $\frac{x + 3}{x - 1} = \frac{1}{3}$ , a student performs the substitution and gets $\frac{-2 + 3}{-2 - 1} = \frac{1}{-3} = -\frac{1}{3}$ . What should the student conclude about this potential solution?
Answer: $x = -2$ is not a solution because $-\frac{1}{3} ≠ \frac{1}{3}$ , so the equation is not satisfied The student's substitution is correct: (-2 + 3)/(-2 - 1) = 1/(-3) = -1/3. Since -1/3 ≠ 1/3, x = -2 does not satisfy the equation and is not a solution. Choice B incorrectly claims 1/3 and -1/3 are equivalent. Choice C incorrectly states both sides equal 1/3. Choice D is wrong because x = -2 makes the denominator -3, not zero (x = 1 would make the denominator zero).
A student solves $2^{x+1} = 8$ and claims $x = 2$ . When checking this solution, which verification process demonstrates the most complete understanding?
Answer: Check that $2^{x+1} = 2^3$ when $x = 2$ , then verify $2^3 = 8$ and confirm uniqueness The most complete verification involves: (1) substituting x = 2 to get $2^{2+1}$ = 2³, (2) confirming 2³ = 8, and (3) recognizing that exponential equations with the same base have unique solutions when the bases equal each other. Choice A is correct but doesn't demonstrate understanding of uniqueness. Choice B incorrectly calculates 2³ as 4 + 1. Choice D shows partial understanding but doesn't clearly demonstrate the exponential relationship.
Maria claims that $x = 3$ is a solution to the equation $2x^2 - 5x - 3 = 0$ . When checking her solution by substitution, which statement best describes what she should conclude?
Answer: The solution is correct because $2(3)^2 - 5(3) - 3 = 18 - 15 - 3 = 0$ To check if x = 3 is a solution, substitute into the original equation: 2(3)² - 5(3) - 3 = 2(9) - 15 - 3 = 18 - 15 - 3 = 0. Since the result equals 0, x = 3 is indeed a solution. Choice B incorrectly calculates 2(3)² as 6 instead of 18. Choice C makes the same error and incorrectly concludes the solution is correct despite getting -9. Choice D incorrectly calculates -5(3) as -8 instead of -15.
A student solves the equation $x^3 - 6x^2 + 11x - 6 = 0$ and claims the solution set is $\{1, 2, 3\}$ . To verify, the student checks $x = 2$ : $2^3 - 6(2^2) + 11(2) - 6 = 8 - 24 + 22 - 6 = 0$ ✓. The student then states the solution set is verified. What is the most significant issue with this verification process?
Answer: All three proposed solutions must be individually verified by substitution to confirm the solution set While the student correctly verified that $x = 2$ is a solution, they must also verify $x = 1$ and $x = 3$ by substitution to confirm the complete solution set. For $x = 1$ : $1 - 6 + 11 - 6 = 0$ ✓. For $x = 3$ : $27 - 54 + 33 - 6 = 0$ ✓. Choice A is incorrect as the arithmetic is right. Choice B misses the point that individual verification is needed. Choice D suggests a solving method rather than addressing the verification issue.
The equation $\frac{2x - 1}{x + 3} = \frac{x + 4}{2x - 5}$ is solved by cross-multiplication, yielding $(2x - 1)(2x - 5) = (x + 4)(x + 3)$ . After expansion and simplification, the result is $3x^2 - 19x - 7 = 0$ . A student finds $x = 7$ as one solution and verifies it by substituting back into the quadratic equation: $3(49) - 19(7) - 7 = 147 - 133 - 7 = 7 \neq 0$ . What should the student conclude?
Answer: The verification method is wrong; the student should substitute $x = 7$ into the original rational equation instead When solving rational equations, it's crucial to understand the difference between verifying solutions in the derived equation versus the original equation. This distinction becomes critical when extraneous solutions might be introduced during the solving process. The student correctly applied cross-multiplication and algebraic manipulation to arrive at the quadratic $3x^2 - 19x - 7 = 0$ . However, when $x = 7$ failed to satisfy this quadratic equation, the student needed to recognize that the verification should occur in the original rational equation, not the derived quadratic. Let's check: substituting $x = 7$ into $\frac{2x - 1}{x + 3} = \frac{x + 4}{2x - 5}$ gives $\frac{13}{10} = \frac{11}{9}$ , which is indeed false, confirming $x = 7$ is not a solution. Option A is wrong because the issue isn't with solving the quadratic—it's with the verification method. Option B incorrectly assumes the expansion was wrong when the algebraic steps were actually correct. Option C suggests restarting entirely, but the student's work up to the verification was sound. Option D correctly identifies that verification should happen in the original equation, where denominators and potential restrictions matter. The key insight is that $x = 7$ likely resulted from an arithmetic error in solving the quadratic, but the student's approach to verification was fundamentally flawed. Study tip: Always verify rational equation solutions by substituting back into the original equation, not derived forms, since the original equation contains the domain restrictions that matter.
A student checks whether $(2, -1)$ is a solution to the system of equations $\begin{cases} 3x + 4y = 2 \\ x - 2y = 4 \end{cases}$ by substitution. After checking both equations, what should the student conclude?
Answer: The point is a solution because it satisfies both equations when properly substituted and calculated To be a solution to a system, the point must satisfy ALL equations. For (2, -1): First equation: 3(2) + 4(-1) = 6 - 4 = 2 ✓. Second equation: 2 - 2(-1) = 2 + 2 = 4 ✓. Since both equations are satisfied, (2, -1) is a solution to the system. Choice A only checks one equation, which is insufficient. Choice B contains a correct calculation but draws the wrong conclusion. Choice D is incorrect as both equations are satisfied.
A quadratic inequality $x^2 - 4x + 3 < 0$ has the proposed solution set $1 < x < 3$ . To verify this solution set, a student tests the boundary values $x = 1$ and $x = 3$ , finding that both make the expression equal zero. The student also tests $x = 2$ and gets $4 - 8 + 3 = -1 < 0$ . What additional verification step is most important?
Answer: Test values outside the proposed interval to confirm the inequality is false there, ensuring the solution set is complete To fully verify a solution set for an inequality, one must confirm that values outside the proposed interval do NOT satisfy the inequality. Testing $x = 0$ or $x = 4$ should yield positive values, confirming the solution set is complete and not missing any intervals. Choice B verifies algebra but not the solution set completeness. Choice C confirms the parabola direction but doesn't verify the solution boundaries. Choice D provides redundant verification within the known interval.
After solving $|2x - 3| = 7$ , a student obtains $x = 5$ and $x = -2$ . When checking these solutions by substitution, what should the student verify to confirm both solutions are correct?
Answer: Check that $|2(5) - 3| = |7| = 7$ and $|2(-2) - 3| = |-7| = 7$ , confirming both solutions For x = 5: |2(5) - 3| = |10 - 3| = |7| = 7 ✓. For x = -2: |2(-2) - 3| = |-4 - 3| = |-7| = 7 ✓. Both solutions are correct. Choice B incorrectly calculates 2(-2) as 4 instead of -4. Choice C shows the correct reasoning but doesn't use absolute value notation consistently. Choice D adds an unnecessary condition about 2x - 3 ≥ 0, which isn't required for absolute value equations.
To verify that $x = 5$ is a solution to $\sqrt{2x - 1} = 3$ , a student substitutes and calculates $\sqrt{2(5) - 1} = \sqrt{9} = 3$ . However, the student should also check whether this solution is valid in the context of the original equation. What additional consideration is most important?
Answer: The student should verify that $2x - 1 ≥ 0$ to ensure the expression under the square root is non-negative When checking solutions involving square roots, it's crucial to verify that the expression under the radical is non-negative. With x = 5: 2(5) - 1 = 9 ≥ 0, so the solution is valid. Choice B mentions extraneous solutions but incorrectly references x² = 25. Choice C is incorrect because √9 = 3 by definition (principal square root). Choice D is unnecessary since the student already verified √9 = 3 directly.
When checking solutions to $x^2 - 4x = 5$ , a student rearranges to $x^2 - 4x - 5 = 0$ and then claims that $x = 5$ and $x = -1$ are solutions. To verify these solutions are correct, what is the most efficient checking method?
Answer: Substitute both values into $x^2 - 4x - 5 = 0$ and verify each gives a result of $0$ When checking solutions to quadratic equations, you want the most direct and efficient verification method that confirms your solutions are correct. The most efficient approach is to substitute both values into the rearranged form $x^2 - 4x - 5 = 0$ and verify each gives zero (choice B). Let's check: For $x = 5$ : $(5)^2 - 4(5) - 5 = 25 - 20 - 5 = 0$ ✓. For $x = -1$ : $(-1)^2 - 4(-1) - 5 = 1 + 4 - 5 = 0$ ✓. This method is straightforward and directly confirms that both values satisfy the equation. Choice A would also work mathematically, but it's less efficient because you need to verify that both sides equal the same non-zero value rather than simply checking for zero. Choice C uses the quadratic formula, which doesn't actually "check" the given solutions—instead, it re-solves the entire problem from scratch, making it inefficient for verification purposes. Choice D suggests factoring to verify, but this approach is backwards; you'd typically factor first to find solutions, not use factoring to check already-found solutions. Study tip: When checking quadratic solutions, always substitute into the form that equals zero. This gives you a simple target (zero) and makes arithmetic errors easier to spot. If your substitution doesn't yield zero, you know immediately that either your solution is wrong or you made a calculation error.
A logarithmic equation $\log_2(x + 3) + \log_2(x - 1) = 3$ has the proposed solution $x = 5$ . During verification, a student computes $\log_2(5 + 3) + \log_2(5 - 1) = \log_2(8) + \log_2(4) = 3 + 2 = 5 \neq 3$ . The student concludes $x = 5$ is not a solution. What error did the student make in the verification process?
Answer: The student should have used the product property: $\log_2(8) + \log_2(4) = \log_2(32) = 5$ , but since $2^3 = 8$ , the answer should be 3 The student made an error in applying logarithm properties. Using the product property: $\log_2(8) + \log_2(4) = \log_2(8 \cdot 4) = \log_2(32)$ . Since $2^5 = 32$ , we have $\log_2(32) = 5 \neq 3$ . However, the student's individual calculations were wrong: $\log_2(8) = 3$ is correct since $2^3 = 8$ , but the sum using the product property gives the same result. The verification correctly shows $x = 5$ is not a solution.
Consider the system of equations: $\begin{cases} 2x + 3y = 12 \\ x - y = 1 \end{cases}$ A student proposes that $(x, y) = (3, 2)$ is the solution. When checking by substitution, which statement best explains the verification process and its implications?
Answer: Both equations must be checked: $2(3) + 3(2) = 12$ ✓ and $3 - 2 = 1$ ✓, confirming the solution satisfies the entire system For a system of equations, a solution must satisfy ALL equations simultaneously. Both substitutions must be performed and verified: First equation: $2(3) + 3(2) = 6 + 6 = 12$ ✓. Second equation: $3 - 2 = 1$ ✓. Since both are true, the solution is verified. Choice A only checks one equation. Choice C misunderstands what the verification should show. Choice D incorrectly assumes checking one equation is sufficient.
The equation $\sqrt{x + 5} = x - 1$ has a proposed solution of $x = 4$ . When checking by substitution: $\sqrt{4 + 5} = \sqrt{9} = 3$ and $4 - 1 = 3$ . Since both sides equal 3, a student concludes the solution is verified. However, the original solving process also yielded $x = 0$ as a potential solution. What should be concluded about the complete solution set?
Answer: Only $x = 4$ is valid; $x = 0$ must be checked by substitution and will likely be an extraneous solution When solving radical equations, squaring both sides can introduce extraneous solutions. While $x = 4$ checks correctly, $x = 0$ must also be verified: $\sqrt{0 + 5} = \sqrt{5} \approx 2.24$ but $0 - 1 = -1$ . Since $2.24 \neq -1$ , $x = 0$ is extraneous. Choice A fails to verify the second solution. Choice C unnecessarily discards the verified solution. Choice D incorrectly describes radical equations.
A student claims that $x = 4$ and $x = -1$ are both solutions to $x^2 - 3x - 4 = 0$ . After checking by substitution, which statement accurately describes the solution set?
Answer: Both values are solutions because $4^2 - 3(4) - 4 = 0$ and $(-1)^2 - 3(-1) - 4 = 0$ Checking x = 4: 4² - 3(4) - 4 = 16 - 12 - 4 = 0 ✓. Checking x = -1: (-1)² - 3(-1) - 4 = 1 + 3 - 4 = 0 ✓. Both values satisfy the equation, so both are solutions. Choice B incorrectly claims the calculation for x = -1 is wrong when it's actually correct. Choice C incorrectly claims the calculation for x = 4 is wrong when it's correct. Choice D incorrectly states neither is a solution.
When verifying that $(1, 4)$ is a solution to $y = x^2 + 3$ , a student substitutes $x = 1$ into the right side and gets $y = 1^2 + 3 = 4$ . Since the $y$ -coordinate of the point is also $4$ , the student concludes the point is on the curve. Is this reasoning complete?
Answer: Yes, the reasoning is complete because substituting $x = 1$ yields $y = 4$ , matching the given point When checking if a point lies on a curve, you're verifying whether the coordinates satisfy the given equation. The equation $y = x^2 + 3$ expresses y as a function of x, meaning for any x-value, there's exactly one corresponding y-value. The student's approach is mathematically sound and complete. By substituting $x = 1$ into the equation, they get $y = 1^2 + 3 = 4$ . Since this matches the y-coordinate of the given point $(1, 4)$ , the point satisfies the equation and lies on the curve. This is the standard method for verifying solutions. Choice A is incorrect because it suggests treating the equation as if both variables are independent, requiring separate verification of $1 = 4^2 + 3$ . This would give $1 = 19$ , which is false, but this check is irrelevant since the equation doesn't claim that x equals $y^2 + 3$ . Choice B suggests an unnecessary additional step. Writing $4 = 1^2 + 3$ is just a different way of expressing the same verification the student already completed correctly. Choice C incorrectly assumes you need to check both $y = x^2 + 3$ and $x = y^2 + 3$ . These are completely different equations describing different curves. You only verify the equation you're given. Choice D correctly recognizes that the student's reasoning is complete. Study tip: For function equations like $y = f(x)$ , substitute the x-coordinate and check if you get the corresponding y-coordinate. Don't overthink it by creating additional equations that weren't given.
An exponential equation $3^{x+1} = 9^{x-2}$ has the proposed solution $x = 5$ . A student verifies by computing $3^{5+1} = 3^6 = 729$ and $9^{5-2} = 9^3 = 729$ . The student concludes the solution is correct. However, another student argues that the verification method is flawed. What is the strongest criticism of this verification approach?
Answer: The verification is mathematically sound; both sides equal 729 when $x = 5$ , confirming the solution The verification is actually correct. $3^6 = 729$ and $9^3 = (3^2)^3 = 3^6 = 729$ , so both sides are equal when $x = 5$ . This confirms the solution regardless of the method used to originally solve it. Choice A is wrong because $3^6 = 729$ and $9^3 = 729$ are correct. Choice B suggests an unnecessary constraint on verification methods. Choice C proposes a preference but doesn't identify a flaw in the current method.
When checking whether $t = 3$ satisfies the inequality $2t - 1 < t + 5$ , a student substitutes and gets $2(3) - 1 < 3 + 5$ , which simplifies to $5 < 8$ . What should the student conclude about $t = 3$ and the solution set?
Answer: $t = 3$ satisfies the inequality, but this alone doesn't determine the complete solution set of the inequality Since 5 < 8 is a true statement, t = 3 does satisfy the inequality. However, checking one value doesn't determine the complete solution set—the student would need to solve the inequality algebraically to find all solutions (t < 6). Choice B incorrectly concludes the solution set is just t = 3. Choice C is wrong because 5 < 8 is true. Choice D makes an unjustified leap about values greater than 3.
To check whether $x = 2$ satisfies $\log_3(x + 7) = 2$ , a student substitutes and calculates $\log_3(2 + 7) = \log_3(9)$ . The student then needs to evaluate $\log_3(9)$ to complete the verification. What should the student find?
Answer: $\log_3(9) = 2$ because $3^2 = 9$ , so $x = 2$ is a valid solution to the equation When checking whether a value satisfies a logarithmic equation, you need to understand what logarithms actually mean. A logarithm asks the question: "What power do I raise the base to get this number?" To verify if $x = 2$ satisfies $\log_3(x + 7) = 2$ , you substitute to get $\log_3(9) = 2$ . Now you need to evaluate $\log_3(9)$ . The key insight is that $\log_3(9)$ asks "what power of 3 gives us 9?" Since $3^2 = 9$ , we have $\log_3(9) = 2$ . This matches the right side of our original equation, confirming that $x = 2$ is indeed a solution. Looking at the wrong answers: Choice A incorrectly states that $9 = 3^3$ , but $3^3 = 27$ , not 9. This is a basic calculation error. Choice C confuses logarithms with division— $\log_3(9)$ is not the same as $\frac{9}{3}$ . Logarithms measure exponents, not quotients. Choice D suggests using a calculator to get a decimal approximation, but this is unnecessary when you can find the exact value using the definition of logarithms. The correct answer is B because $3^2 = 9$ , so $\log_3(9) = 2$ , confirming our solution. Remember: always convert logarithms back to exponential form when possible. If $\log_b(y) = x$ , then $b^x = y$ . This direct relationship will help you evaluate logarithms quickly and avoid calculator dependency on exams.
A student claims that $x = 0$ and $x = 3$ are solutions to $x(x - 3) = 0$ . After checking by substitution, the student should conclude:
Answer: Both solutions are correct: $0(0 - 3) = 0(-3) = 0$ and $3(3 - 3) = 3(0) = 0$ When you encounter an equation like $x(x - 3) = 0$ , you're dealing with the zero product property, which states that if two factors multiply to equal zero, then at least one of the factors must be zero. To verify solutions, you must substitute each proposed value back into the original equation and check if both sides are equal. Let's check $x = 0$ : Substituting into $x(x - 3) = 0$ gives us $0(0 - 3) = 0(-3) = 0$ . Since any number multiplied by zero equals zero, this gives us $0 = 0$ , which is true. Now let's check $x = 3$ : Substituting gives us $3(3 - 3) = 3(0) = 0$ . Again, since anything multiplied by zero equals zero, we get $0 = 0$ , which is also true. Both solutions are correct, confirming answer D. Answer A incorrectly calculates both substitutions. It suggests $0(0 - 3) = -3$ , which ignores that zero times anything equals zero, and $3(3 - 3) = 9$ , which incorrectly computes $3 \times 0$ . Answer B correctly verifies $x = 0$ but wrongly claims $3(3 - 3) = 3$ , failing to recognize that $3 - 3 = 0$ and $3 \times 0 = 0$ . Answer C makes the opposite error: correctly verifying $x = 3$ but incorrectly stating that $0(-3) = -3$ , forgetting that zero times any number equals zero. Study tip: Always remember that zero multiplied by any number equals zero. When checking solutions to factored equations, substitute carefully and apply the zero product property systematically.

Start this quiz