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A2B7-E455
wer.
xist for 
|(1)/(2)x+1|=5 ?
ights reserved.
om/assessments-delivery/print/a/45552884

A22B77-E455455\newlinewer.\newlinexist for 12x+1=5 \left|\frac{1}{2} x+1\right|=5 ?\newlineights reserved.\newlineom/assessments-delivery/print/a/4555288445552884

Full solution

Q. A22B77-E455455\newlinewer.\newlinexist for 12x+1=5 \left|\frac{1}{2} x+1\right|=5 ?\newlineights reserved.\newlineom/assessments-delivery/print/a/4555288445552884
  1. Isolate absolute value: Isolate the absolute value expression. 12x+1=5|\frac{1}{2}x + 1| = 5
  2. Consider positive case: Consider the positive case of the absolute value. \newline(12)x+1=5(\frac{1}{2})x + 1 = 5
  3. Subtract and simplify: Subtract 11 from both sides.\newline(12)x=51(\frac{1}{2})x = 5 - 1\newline(12)x=4(\frac{1}{2})x = 4
  4. Multiply to solve: Multiply both sides by 22 to solve for xx.x=4×2x = 4 \times 2x=8x = 8
  5. Consider negative case: Now consider the negative case of the absolute value. \newline12x+1=5\frac{1}{2}x + 1 = -5

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