Solve absolute value inequality. |2(x−1)+4|≤8

Question

Solve absolute value inequality.  |2(x−1)+4|≤8

Leonard Stokes · Accepted Answer

Step 1
We have to solve the absolute value inequality:
$| 2 (x - 1) + 4 | \leq 8$
We know for any modulus function,
$| f (x) | = {\begin{cases} f (x) & i f f (x) \geq 0 \\ - f (x) & i f f (x) < 0 \end{cases}$
We also know that if $| f (x) | \leq$ a then it must satisfy that
$- a \leq f (x) \leq a$
Applying above condition for given inequality,
$- a \leq f (x) \leq a$
$- 8 \leq 2 (x - 1) + 4 \leq 8$
Step 2
Simplifying further,
$- 8 - 4 \leq 2 (x - 1) + 4 - 4 \leq 8 - 4$ (substracted 4)
$- 12 \leq 2 (x - 1) \leq 4$
$\frac{- 12}{2} \leq \frac{2 (x - 1)}{2} \leq \frac{4}{2}$ (divided all sides by 2)
$- 6 \leq (x - 1) \leq 2$
$- 6 + 1 \leq x - 1 + 1 \leq 2 + 1$
$- 5 \leq x \leq 3$
In interval notation we can write it as $x \in [- 5, 3]$ .
Hence, solution of the absolute value inequality is $x \in [- 5, 3]$ .

Solve absolute value inequality. |2(x-1)+4|\leq8

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