Math 1 Quiz: Checking Solutions By Substitution

What this quiz covers

This quiz focuses on Checking Solutions By Substitution, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The student should verify that $2x - 1 ≥ 0$ to ensure the expression under the square root is non-negative (correct answer)
2. The student should check if $x = 5$ produces any extraneous solutions by substituting back into $x^2 = 25$
3. The student should confirm that $\sqrt{9}$ equals $3$ and not $-3$ since square roots are always positive
4. The student should verify that both sides of the equation are equal when squared: $(\sqrt{9})^2 = 3^2$

Question 2

1. No, the student should verify by substituting both coordinates: $4 = 1^2 + 3$ and $1 = 4^2 + 3$
2. No, the student should also substitute $y = 4$ into the equation to verify: $4 = 1^2 + 3$
3. No, the student should check that the point satisfies $x = y^2 + 3$ as well as $y = x^2 + 3$
4. Yes, the reasoning is complete because substituting $x = 1$ yields $y = 4$, matching the given point (correct answer)

Question 3

1. The solution is correct, but the student should also verify there are no other solutions by checking $x = -1$
2. The solution is incorrect because $\sqrt{9} = ±3$, and the student must consider both positive and negative roots
3. The solution needs verification by checking the domain restriction $x + 5 ≥ 0$ and $x - 1 ≥ 0$ for radical equations
4. The solution $x = 4$ is correct since both sides equal $3$ when substituted into the original equation (correct answer)

Question 4

1. Substitute both values into the original form $x^2 - 4x = 5$ and check if both sides are equal
2. Substitute both values into $x^2 - 4x - 5 = 0$ and verify each gives a result of $0$ (correct answer)
3. Use the quadratic formula with $a = 1$, $b = -4$, $c = -5$ to confirm the solutions algebraically
4. Factor $x^2 - 4x - 5$ as $(x - 5)(x + 1)$ and verify this equals the original expression

Question 5

1. $\log_3(9) = 3$ because $9 = 3^3$, so $x = 2$ is not a solution to the equation
2. $\log_3(9) = 2$ because $3^2 = 9$, so $x = 2$ is a valid solution to the equation (correct answer)
3. $\log_3(9) = \frac{9}{3} = 3$, so $x = 2$ is not a solution to the original logarithmic equation
4. $\log_3(9) = 2.197$ using a calculator, which approximately equals $2$, so $x = 2$ is a solution

Question 6

1. $x = -2$ is not a solution because $-\frac{1}{3} ≠ \frac{1}{3}$, so the equation is not satisfied (correct answer)
2. $x = -2$ is a solution because the fractions $\frac{1}{3}$ and $-\frac{1}{3}$ are equivalent when simplified
3. $x = -2$ is a solution because both sides equal $\frac{1}{3}$ when the substitution is performed correctly
4. $x = -2$ is not a solution because it makes the denominator zero in the original equation

Question 7

1. $t = 3$ satisfies the inequality, but this alone doesn't determine the complete solution set of the inequality (correct answer)
2. $t = 3$ satisfies the inequality, and since $5 < 8$ is true, the solution set is $t = 3$
3. $t = 3$ does not satisfy the inequality because $5 < 8$ should be written as $8 > 5$
4. $t = 3$ satisfies the inequality, proving that all values greater than $3$ are solutions to this inequality

Question 8

1. The point is a solution because it satisfies the first equation: $3(2) + 4(-1) = 6 - 4 = 2$
2. The point is not a solution because it fails the second equation: $2 - 2(-1) = 2 + 2 = 4$
3. The point is a solution because it satisfies both equations when properly substituted and calculated (correct answer)
4. The point is not a solution because it fails both equations when the calculations are performed correctly

Question 9

1. The solution $x = 7$ is incorrect, and the student should re-solve the quadratic equation
2. The quadratic expansion was incorrect since the verification failed, indicating an algebraic error earlier
3. Both the solution and verification contain errors; the student should restart the entire problem
4. The verification method is wrong; the student should substitute $x = 7$ into the original rational equation instead (correct answer)

Question 10

1. Test values outside the proposed interval to confirm the inequality is false there, ensuring the solution set is complete (correct answer)
2. Verify the factorization $x^2 - 4x + 3 = (x-1)(x-3)$ by expanding to ensure no algebraic errors occurred
3. Check that the parabola opens upward by confirming the coefficient of $x^2$ is positive, validating the interval direction
4. Substitute additional points within the interval such as $x = 1.5$ and $x = 2.5$ to verify consistency throughout

Question 11

1. Both $x = 4$ and $x = 0$ are valid since they both came from the algebraic solving process
2. Only $x = 4$ is valid; $x = 0$ must be checked by substitution and will likely be an extraneous solution (correct answer)
3. Neither solution can be trusted without re-solving the equation using a different algebraic method
4. The solution set is incomplete because radical equations typically have infinitely many solutions

Question 12

1. The arithmetic calculation contains an error that invalidates the verification of $x = 2$
2. A cubic equation can have at most 3 solutions, so finding one correct solution doesn't verify the complete solution set
3. All three proposed solutions must be individually verified by substitution to confirm the solution set (correct answer)
4. The student should have used synthetic division rather than direct substitution to verify cubic solutions

Question 13

1. The student incorrectly calculated $\log_2(8) = 3$ when it should be $\log_2(8) = 2$
2. The student should have used the product property: $\log_2(8) + \log_2(4) = \log_2(32) = 5$, confirming $x = 5$ is indeed not a solution
3. 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 (correct answer)
4. The student failed to check the domain restriction that $x > 1$ before performing the logarithmic calculations

Question 14

1. The calculations contain arithmetic errors that invalidate the verification process entirely
2. Exponential equations should be verified using logarithms, not direct computation with large numbers
3. The student should have converted both sides to the same base before substituting to avoid computational errors
4. The verification is mathematically sound; both sides equal 729 when $x = 5$, confirming the solution (correct answer)

Question 15

1. Both solutions are valid because they both satisfy the original equation when substituted (correct answer)
2. Only $x = 5$ is valid because it produces a positive expression inside the absolute value bars
3. Only $x = -2$ is valid because absolute value equations typically have negative solutions
4. Neither solution is complete without checking the corresponding negative case $|2x - 3| = -7$

Question 16

1. Substituting into the first equation gives $2(3) + 3(2) = 12$, which is true, so the solution is verified completely
2. Both equations must be checked: $2(3) + 3(2) = 12$ ✓ and $3 - 2 = 1$ ✓, confirming the solution satisfies the entire system (correct answer)
3. The solution is invalid because substituting gives $2(3) + 3(2) = 12$ and $3 - 2 = 1$, but these don't match the original coefficients
4. Only one equation needs verification since systems have unique solutions, and $2(3) + 3(2) = 12$ confirms correctness

Question 17

1. Direct substitution gives $\frac{0}{0}$, which is indeterminate; the student should factor and simplify before substituting (correct answer)
2. Direct substitution gives $\frac{1}{0}$, which is undefined; therefore $x = 1$ cannot be a solution to this equation
3. Direct substitution works fine: $\frac{1 - 1}{1 - 1} = \frac{0}{0} = 0$ and $1 + 1 = 2$, so $x = 1$ is not a solution
4. Direct substitution gives $\frac{2}{0}$, which means $x = 1$ makes the equation undefined and invalid

Question 18

1. The solution is correct because $2(3)^2 - 5(3) - 3 = 18 - 15 - 3 = 0$ (correct answer)
2. The solution is incorrect because $2(3)^2 - 5(3) - 3 = 6 - 15 - 3 = -12 ≠ 0$
3. The solution is correct because $2(3)^2 - 5(3) - 3 = 9 - 15 - 3 = -9 ≠ 0$
4. The solution is incorrect because $2(3)^2 - 5(3) - 3 = 18 - 8 - 3 = 7 ≠ 0$

Question 19

1. Neither solution is correct: $0(0 - 3) = -3$ and $3(3 - 3) = 9$ when calculated properly
2. Only $x = 0$ is correct: $0(0 - 3) = 0$, but $3(3 - 3) = 3(0) = 3 ≠ 0$
3. Only $x = 3$ is correct: $3(3 - 3) = 0$, but $0(0 - 3) = 0(-3) = -3 ≠ 0$
4. Both solutions are correct: $0(0 - 3) = 0(-3) = 0$ and $3(3 - 3) = 3(0) = 0$ (correct answer)

Question 20

1. $x = -3$ gives $\frac{1}{0} + \frac{1}{-5} = 0$, which is undefined, but $x = 2$ gives $\frac{1}{5} + \frac{1}{0} = 0$, which works
2. $x = -3$ makes the first fraction undefined, and $x = 2$ makes the second fraction undefined, so neither can be solutions (correct answer)
3. Both values can be substituted successfully: $\frac{1}{0} + \frac{1}{-5} = 0$ and $\frac{1}{5} + \frac{1}{0} = 0$
4. $x = -3$ gives $\frac{1}{0} + \frac{1}{-5}$ and $x = 2$ gives $\frac{1}{5} + \frac{1}{0}$, both resulting in undefined expressions