Question

Difficulty: HardLinear Equations in One and Two Variables

If xx satisfies the equation ∣8x+13∣=10x+5|8x + 13| = 10x + 5, what is the sum of all valid real solutions for xx?

  1. 4Answer
  2. B
    3
  3. C
    -1
  4. D
    -4
  5. E
    5

Answer

The sum of all valid real solutions for xx is 4.
The solution x=4x = 4 is the only value of xx that satisfies ∣8x+13∣=10x+5|8x + 13| = 10x + 5 because evaluating both sides yields 45=4545 = 45. The alternative algebraic candidate, x=−1x = -1, produces a right-hand side of −5-5, which is impossible for an absolute value expression. Thus, the sum of all valid real solutions is simply 4.

Step-by-Step Solution

1
Split the absolute value equation into two linear case equations
Case 1: 8x+13=10x+58x + 13 = 10x + 5; Case 2: 8x+13=−(10x+5)8x + 13 = -(10x + 5)
By definition of absolute value, ∣u∣=v|u| = v implies u=vu = v or u=−vu = -v, provided v≥0v \ge 0.
2
Solve Case 1 for xx
2x=8  ⟹  x=42x = 8 \implies x = 4
Subtract 8x8x and 55 from both sides to isolate xx.
3
Solve Case 2 for xx
8x+13=−10x−5  ⟹  18x=−18  ⟹  x=−18x + 13 = -10x - 5 \implies 18x = -18 \implies x = -1
Expand the right-hand side and combine like terms to isolate xx.
4
Test candidate solutions in the original equation to eliminate extraneous roots
For x=4x = 4: ∣8(4)+13∣=∣45∣=45|8(4)+13| = |45| = 45 and 10(4)+5=4510(4)+5 = 45 (Valid). For x=−1x = -1: ∣8(−1)+13∣=∣5∣=5|8(-1)+13| = |5| = 5 and 10(−1)+5=−510(-1)+5 = -5 (Extraneous).
An absolute value output cannot equal a negative number.
5
Calculate the sum of all valid solutions
The only valid solution is x=4x = 4, so the sum is 44.
Only valid roots must be summed.

Key Concept

Absolute Value Equations and Checking for Extraneous Solutions
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