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Zorluk: KolayInequalities and Absolute Value Equations

What is the real solution to the equation ∣x−5∣=2x+1|x - 5| = 2x + 1?

  1. x=43x = \frac{4}{3}Cevap
  2. B
    x=−6x = -6
  3. C
    x=−6x = -6 and x=43x = \frac{4}{3}
  4. D
    x=−43x = -\frac{4}{3}
  5. E
    x=6x = 6

Cevap

x=43x = \frac{4}{3}
The correct answer is x=43x = \frac{4}{3}. Setting x−5=−(2x+1)x - 5 = -(2x + 1) gives x−5=−2x−1x - 5 = -2x - 1, which solves to 3x=43x = 4 or x=43x = \frac{4}{3}. Plugging x=43x = \frac{4}{3} into both sides of the original equation yields 113=113\frac{11}{3} = \frac{11}{3}, satisfying the equation.

Adım Adım Çözüm

1
Set up the two cases for the absolute value equation
Case 1: x−5=2x+1x - 5 = 2x + 1; Case 2: x−5=−(2x+1)x - 5 = -(2x + 1)
By definition, ∣A∣=B|A| = B implies A=BA = B or A=−BA = -B, provided B≥0B \geq 0.
2
Solve Case 1 algebraically
x−5=2x+1  ⟹  x=−6x - 5 = 2x + 1 \implies x = -6
Subtracting xx and 11 from both sides isolates xx.
3
Solve Case 2 algebraically
x−5=−2x−1  ⟹  3x=4  ⟹  x=43x - 5 = -2x - 1 \implies 3x = 4 \implies x = \frac{4}{3}
Distributing the negative sign and adding 2x2x and 55 to both sides isolates xx.
4
Check for extraneous solutions in the original equation ∣x−5∣=2x+1|x - 5| = 2x + 1
For x=−6x = -6: ∣−6−5∣=∣−11∣=11|-6 - 5| = |-11| = 11, but 2(−6)+1=−11≠112(-6) + 1 = -11 \neq 11 (extraneous). For x=43x = \frac{4}{3}: ∣43−5∣=∣−113∣=113|\frac{4}{3} - 5| = |-\frac{11}{3}| = \frac{11}{3}, and 2(43)+1=1132(\frac{4}{3}) + 1 = \frac{11}{3} (valid).
An absolute value expression cannot equal a negative number.

Anahtar Kavram

Solving absolute value equations with a variable on the right-hand side requires checking for extraneous solutions.
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