Part II Review

1: B. Gather the x-terms: 6x−3x=16+5, so 3x=21, giving x=7. Choice D reverses the sign of the constants while combining, computing 3x=−21 instead of 3x=21.

2: J. Subtract 9 from both sides: −4x≤−12. Dividing by −4 reverses the inequality: x≥3. Choice F keeps the inequality direction unchanged after dividing by −4.

3: D. From the second equation, x=y+1; substituting gives 5(y+1)+2y=19, so 7y=14, y=2, x=3, and x+y=5. Choice A copies the constant from the second equation directly instead of computing x + y.

4: G. Divide both sides by lw: h = V ÷ (lw). Choice F inverts the fraction, placing V in the denominator instead of the numerator.

5: B. For no solution, the coefficient ratios must match while the constant ratios differ: 3/1=k/2, giving k=6; checking, 9/6=1.5 does not equal 3/1, confirming no solution. Choice A copies the x-coefficient of the second equation directly instead of solving the ratio equation.

6: G. Factor: (x+5)(x−2)=0, giving x=−5 or x=2. Choice F reverses the sign of both roots.

7: A. Using the quadratic formula: discriminant=25+24=49, √49=7. x=(5±7)/6, giving x=2 or x=−1/3. Choice B reverses the sign of both roots.

8: G. Discriminant: (2)²−4(1)(9)=4−36=−32, which is negative, giving two complex solutions. Choice F reports the discriminant with the sign reversed, along with the real-solution conclusion a positive discriminant would produce.

9: A. x-coordinate: −(−10)/2(1)=5. y-coordinate: 5²−10(5)+21=25−50+21=−4. Choice B reverses the sign of the x-coordinate only.

10: F. One real solution requires the discriminant to equal 0: k²−144=0, so k=12 or k=−12. Choice G reports only the positive root, dropping the negative solution.

11: D. Testing x=2 shows it is a root; dividing gives x²−x−6=(x−3)(x+2). The full factorization is (x−2)(x−3)(x+2)=0, giving x=2, 3, or −2. Choice A reverses the sign of the root x=3 only.

12: J. Rewrite 125 as a power of 5: 5^(x+1)=5³. Set the exponents equal: x+1=3, giving x=2. Choice F reports the exponent value 3 instead of solving for x.

13: C. Substitute x=−2 and set the result equal to 0: −8+4k−10+6=0, so 4k−12=0, giving k=3. Choice A drops the negative sign, reporting the sign-reversed value of k.

14: H. Rewrite 8 as a power of 2: 2^(2x)=2^(3x−3). Set the exponents equal: 2x=3x−3, giving x=3. Choice F makes a sign error while solving 2x=3x−3, giving x=−3 instead of x=3.

15: C. Rewrite 81 as a power of 3: 3^(x−1)=3⁴. Set the exponents equal: x−1=4, giving x=5. Then y=x²−1=25−1=24. Choice A subtracts 2 instead of 1 from x², computing 25−2 instead of 25−1.

16: F. Case 1: 4x−3=13→x=4. Case 2: 4x−3=−13→x=−2.5. Choice G reverses the sign of both solutions.

17: A. Rewrite as −9<x+6<9, then subtract 6 from every part: −15<x<3. Choice D treats the expression as x−6 inside the absolute value instead of x+6, shifting the boundaries the wrong direction.

18: H. Case 1: 2x+5≥11→x≥3. Case 2: 2x+5≤−11→x≤−8. Choice F reverses both inequality directions.

19: B. Case 2 gives x=2/3, and 2(2/3)+1=7/3≥0 checks. Case 1 gives x=−4, but 2(−4)+1=−7<0, impossible for an absolute value; rejected. Choice A reports the extraneous solution from the case that fails the nonnegative check.

20: F. Case 1: 2x−1=x+7→x=8; check: |15|=15 and 8+7=15, matches. Case 2: 2x−1=−(x+7)→x=−2; check: |−5|=5 and −2+7=5, also matches. Both cases are valid here. Choice G reports only one of the two solutions, missing that both cases check out in the original equation.