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Difficulty: HardLinear Equations in One and Two Variables

If xx is a real number that satisfies the equation ∣x+7∣=2x+3|x + 7| = 2x + 3, what is the sum of all valid real solutions for xx?

  1. A
    23\frac{2}{3}
  2. 44Answer
  3. C
    −103-\frac{10}{3}
  4. D
    −23-\frac{2}{3}
  5. E
    223\frac{22}{3}

Answer

The sum of all valid real solutions is 4.
To solve ∣x+7∣=2x+3|x + 7| = 2x + 3, we break the absolute value into two linear cases: x+7=2x+3x + 7 = 2x + 3 giving x=4x = 4, and x+7=−(2x+3)x + 7 = -(2x + 3) giving x=−103x = -\frac{10}{3}. We must test both candidate solutions in the original equation because absolute value outputs must be non-negative. Testing x=4x = 4 gives ∣11∣=11|11| = 11, which is true. Testing x=−103x = -\frac{10}{3} gives ∣113∣=−113|\frac{11}{3}| = -\frac{11}{3}, which is false because an absolute value cannot equal a negative number. Thus, x=4x = 4 is the only valid solution, making the sum of all valid solutions equal to 44.

Step-by-Step Solution

1
Set up the two linear cases for the absolute value equation ∣x+7∣=2x+3|x + 7| = 2x + 3.
Case 1: x+7=2x+3x + 7 = 2x + 3; Case 2: x+7=−(2x+3)x + 7 = -(2x + 3).
By definition, ∣A∣=B|A| = B implies A=BA = B or A=−BA = -B, provided B≥0B \ge 0.
2
Solve Case 1 for xx.
x+7=2x+3  ⟹  2x−x=7−3  ⟹  x=4x + 7 = 2x + 3 \implies 2x - x = 7 - 3 \implies x = 4.
Isolating xx gives the first candidate solution.
3
Solve Case 2 for xx.
x+7=−2x−3  ⟹  3x=−10  ⟹  x=−103x + 7 = -2x - 3 \implies 3x = -10 \implies x = -\frac{10}{3}.
Expanding the negative sign and isolating xx gives the second candidate solution.
4
Check candidate solutions in the original equation ∣x+7∣=2x+3|x + 7| = 2x + 3 to eliminate extraneous solutions.
For x=4x = 4: ∣4+7∣=11|4 + 7| = 11 and 2(4)+3=112(4) + 3 = 11 (Valid).
For x=−103x = -\frac{10}{3}: ∣−103+7∣=113|-\frac{10}{3} + 7| = \frac{11}{3}, but 2(−103)+3=−1132(-\frac{10}{3}) + 3 = -\frac{11}{3} (Extraneous, since 113≠−113\frac{11}{3} \neq -\frac{11}{3}).
An absolute value expression cannot equal a negative number, so candidate values resulting in a negative right side are invalid.
5
Calculate the sum of all valid real solutions.
Sum = 44.
Since x=4x = 4 is the only valid solution, the sum is simply 44.

Key Concept

Solving Absolute Value Linear Equations and Validating against Extraneous Solutions
Estimated Time:2m 0s
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Topics

Linear Equations in One and Two Variables

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