Solve an absolute-value equation by splitting into its two sign cases, then check each candidate against the equation. The bars around 2x − 20 mean the quantity inside could be positive or negative, so write two cases.
Case 1 (inside is positive or zero): 2x − 20 = x − 4. Subtract x from both sides: x − 20 = −4, so x = 16. Check: |2(16) − 20| = |12| = 12, and 16 − 4 = 12. The two sides match, so x = 16 is valid.
Case 2 (inside is negative): 2x − 20 = −(x − 4) = −x + 4. Add x to both sides: 3x − 20 = 4, so 3x = 24 and x = 8. Check: |2(8) − 20| = |−4| = 4, and 8 − 4 = 4. The two sides match, so x = 8 is valid.
both cases give a genuine solution, so the truck is flagged at x = 16 and at x = 8. The sum is 16 + 8 = 24, which is (D).
Why the wrong choices are tempting: (C) 16 keeps only the positive case and forgets the negative case, a very common slip on absolute-value equations. (A) 8 makes the mirror-image mistake, keeping only the negative case. (B) 10 comes from setting the inside of the absolute value equal to zero (2x − 20 = 0); that is the point where the expression changes sign, not a solution to the equation. (E) 32 takes the positive-case root 16 and doubles it on the false idea that the answer must come in a symmetric pair of equal size. The only choice that correctly handles both sign cases and adds the two valid roots is (D) 24.