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solve each inequality graph 3. ( 2|x| leq 6 ) ( \frac{1}{2}|x| + 2 < 3 )

Question

solve each inequality graph

  1. ( 2|x| leq 6 )

( \frac{1}{2}|x| + 2 < 3 )

Explanation:

Step1: Isolate absolute value term

Divide both sides by 2:
$\frac{2|x|}{2} \leq \frac{6}{2}$
Simplify to get $|x| \leq 3$

Step2: Rewrite without absolute value

For $|x| \leq a$ ($a>0$), $-a \leq x \leq a$:
$-3 \leq x \leq 3$

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Step1: Isolate absolute value term

Subtract 2 from both sides:
$\frac{1}{2}|x| + 2 - 2 < 3 - 2$
Simplify to get $\frac{1}{2}|x| < 1$

Step2: Eliminate coefficient of absolute value

Multiply both sides by 2:
$2 \times \frac{1}{2}|x| < 2 \times 1$
Simplify to get $|x| < 2$

Step3: Rewrite without absolute value

For $|x| < a$ ($a>0$), $-a < x < a$:
$-2 < x < 2$

Answer:

  1. For $2|x| \leq 6$: $-3 \leq x \leq 3$ (on the graph, this is a closed interval from $x=-3$ to $x=3$, with filled dots at $-3$ and $3$ and the segment between them shaded)
  2. For $\frac{1}{2}|x| + 2 < 3$: $-2 < x < 2$ (on the graph, this is an open interval from $x=-2$ to $x=2$, with open dots at $-2$ and $2$ and the segment between them shaded)