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Difficulty: Very hardLinear Equations in One and Two Variables

What is the sum of all real solutions to the absolute value equation ∣2x−3∣+∣x+5∣=12−x|2x - 3| + |x + 5| = 12 - x?

  1. −92-\frac{9}{2}Answer
  2. B
    52\frac{5}{2}
  3. C
    −12-\frac{1}{2}
  4. D
    92\frac{9}{2}
  5. E
    192\frac{19}{2}

Answer

The sum of all real solutions to the equation is −92-\frac{9}{2}.
By splitting the real number line into three intervals based on the critical points x=−5x = -5 and x=32x = \frac{3}{2}, we find two valid solutions: x=52x = \frac{5}{2} (from x≥32x \ge \frac{3}{2}) and x=−7x = -7 (from x<−5x < -5). Their sum is 52+(−7)=−92\frac{5}{2} + (-7) = -\frac{9}{2}.

Step-by-Step Solution

1
Identify critical points for the absolute value expressions
The critical points are x=32x = \frac{3}{2} and x=−5x = -5.
The expressions inside the absolute values, 2x−32x - 3 and x+5x + 5, change signs at x=32x = \frac{3}{2} and x=−5x = -5 respectively.
2
Evaluate Region 1 (x≥32x \ge \frac{3}{2})
(2x−3)+(x+5)=12−x  ⟹  3x+2=12−x  ⟹  4x=10  ⟹  x=52(2x - 3) + (x + 5) = 12 - x \implies 3x + 2 = 12 - x \implies 4x = 10 \implies x = \frac{5}{2}.
In this region, both 2x−3≥02x - 3 \ge 0 and x+5>0x + 5 > 0, so absolute value bars can be removed directly. Since 52≥32\frac{5}{2} \ge \frac{3}{2}, x=52x = \frac{5}{2} is a valid solution.
3
Evaluate Region 2 (−5≤x<32-5 \le x < \frac{3}{2})
−(2x−3)+(x+5)=12−x  ⟹  −x+8=12−x  ⟹  8=12-(2x - 3) + (x + 5) = 12 - x \implies -x + 8 = 12 - x \implies 8 = 12 (No solution).
In this interval, 2x−3<02x - 3 < 0 while x+5≥0x + 5 \ge 0. The resulting equation produces a contradiction, so there are no solutions in this interval.
4
Evaluate Region 3 (x<−5x < -5)
−(2x−3)−(x+5)=12−x  ⟹  −3x−2=12−x  ⟹  −2x=14  ⟹  x=−7-(2x - 3) - (x + 5) = 12 - x \implies -3x - 2 = 12 - x \implies -2x = 14 \implies x = -7.
In this region, both 2x−3<02x - 3 < 0 and x+5<0x + 5 < 0. Since −7<−5-7 < -5, x=−7x = -7 is a valid solution.
5
Calculate the sum of all valid solutions
52+(−7)=52−142=−92\frac{5}{2} + (-7) = \frac{5}{2} - \frac{14}{2} = -\frac{9}{2}.
Combining the valid roots from Region 1 and Region 3 gives the final requested sum.

Key Concept

Solving piecewise linear equations involving multiple absolute value terms.
Estimated Time:2m 30s
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Topics

Linear Equations in One and Two Variables

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