Question

1. $-|x| + 5 geq -1$
2. $-\frac{1}{2}x + 2 > |x - 1|$
3. $2|x| + 3 leq -x + 6$

Explanation:

Step1: Solve inequality 4: Isolate |x|

$-|x| + 5 \geq -1$
Subtract 5 from both sides:
$-|x| \geq -6$
Multiply by -1 (reverse inequality):
$|x| \leq 6$
This translates to:
$-6 \leq x \leq 6$

Step2: Solve inequality 5: Split absolute value

$-\frac{1}{2}x + 2 > |x - 1|$
This requires two cases:

Case 1: $x - 1 \geq 0$ (i.e., $x \geq 1$)

$-\frac{1}{2}x + 2 > x - 1$
Add $\frac{1}{2}x$ to both sides:
$2 > \frac{3}{2}x - 1$
Add 1 to both sides:
$3 > \frac{3}{2}x$
Multiply by $\frac{2}{3}$:
$2 > x$
Combine with $x \geq 1$: $1 \leq x < 2$

Case 2: $x - 1 < 0$ (i.e., $x < 1$)

$-\frac{1}{2}x + 2 > -(x - 1)$
Simplify right side:
$-\frac{1}{2}x + 2 > -x + 1$
Add $x$ to both sides:
$\frac{1}{2}x + 2 > 1$
Subtract 2 from both sides:
$\frac{1}{2}x > -1$
Multiply by 2:
$x > -2$
Combine with $x < 1$: $-2 < x < 1$

Combine both cases: $-2 < x < 2$

Step3: Solve inequality 6: Isolate |x|

$2|x| + 3 \leq -x + 6$
Subtract 3 from both sides:
$2|x| \leq -x + 3$
This requires two cases:

Case 1: $x \geq 0$

$2x \leq -x + 3$
Add $x$ to both sides:
$3x \leq 3$
Divide by 3:
$x \leq 1$
Combine with $x \geq 0$: $0 \leq x \leq 1$

Case 2: $x < 0$

$2(-x) \leq -x + 3$
Simplify left side:
$-2x \leq -x + 3$
Add $2x$ to both sides:
$0 \leq x + 3$
Subtract 3 from both sides:
$-3 \leq x$
Combine with $x < 0$: $-3 \leq x < 0$

Combine both cases: $-3 \leq x \leq 1$

Answer:

1. $-6 \leq x \leq 6$
2. $-2 < x < 2$
3. $-3 \leq x \leq 1$