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Difficulty: HardLinear Equations in One and Two Variables

If xx is a real number that satisfies the equation ∣5−2x∣=3x−10|5 - 2x| = 3x - 10, what is the value of ∣x−4∣+2x|x - 4| + 2x?

  1. A
    5
  2. B
    7
  3. 11Answer
  4. D
    14
  5. E
    18

Answer

11
Solving the equation ∣5−2x∣=3x−10|5 - 2x| = 3x - 10 yields two potential values: x=3x = 3 and x=5x = 5. Substituting x=3x = 3 into the original equation yields ∣−1∣=−1|-1| = -1, which is impossible because an absolute value cannot be negative. Therefore, x=3x = 3 is an extraneous root. Substituting x=5x = 5 yields ∣−5∣=5|-5| = 5, which is true. Evaluating the requested expression ∣x−4∣+2x|x - 4| + 2x at x=5x = 5 gives ∣5−4∣+2(5)=1+10=11|5 - 4| + 2(5) = 1 + 10 = 11.

Step-by-Step Solution

1
Set up the two algebraic cases for the absolute value equation ∣5−2x∣=3x−10|5 - 2x| = 3x - 10.
Case 1: 5−2x=3x−105 - 2x = 3x - 10 when 5−2x≥05 - 2x \ge 0 (x≤2.5x \le 2.5).
Case 2: −(5−2x)=3x−10-(5 - 2x) = 3x - 10 when 5−2x<05 - 2x < 0 (x>2.5x > 2.5).
An absolute value expression ∣u∣|u| equals uu when u≥0u \ge 0 and −u-u when u<0u < 0.
2
Solve Case 1 algebraically.
5−2x=3x−10  ⟹  15=5x  ⟹  x=35 - 2x = 3x - 10 \implies 15 = 5x \implies x = 3.
Isolate the variable xx on one side of the equation.
3
Solve Case 2 algebraically.
2x - 5 = 3x - 10 \implies 5 = x \implies x = 5$.
Simplify and isolate xx.
4
Check both candidate solutions in the original equation to filter out extraneous roots.
For x=3x = 3: ∣5−2(3)∣=3(3)−10  ⟹  ∣−1∣=−1  ⟹  1=−1|5 - 2(3)| = 3(3) - 10 \implies |-1| = -1 \implies 1 = -1 (False, extraneous).
For x=5x = 5: ∣5−2(5)∣=3(5)−10  ⟹  ∣−5∣=5  ⟹  5=5|5 - 2(5)| = 3(5) - 10 \implies |-5| = 5 \implies 5 = 5 (True, valid).
Since the right-hand side 3x−103x - 10 must be non-negative for the absolute value to hold, candidate solutions must be tested in the original equation.
5
Substitute the valid solution x=5x = 5 into the targeted expression ∣x−4∣+2x|x - 4| + 2x.
∣5−4∣+2(5)=∣1∣+10=1+10=11|5 - 4| + 2(5) = |1| + 10 = 1 + 10 = 11.
Evaluate the target expression using the single real value x=5x = 5 that satisfies the given linear absolute value equation.

Key Concept

Solving absolute value linear equations and identifying extraneous solutions
Estimated Time:2m 0s
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Topics

Linear Equations in One and Two Variables

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