# Solve absolute value inequality. 4+|3-x/3|\geq 9

Solve absolute value inequality.
$4+|3-\frac{x}{3}|\ge 9$
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stomachdm
Step 1
Given inequality ,
$4+|3-\frac{x}{3}|\ge 9$
$|3-\frac{x}{3}|\ge 9-4$
$|3-\frac{x}{3}|\ge 5$
After removing modulus function, we get two cases.
$3-\frac{x}{3}\ge 5$...(1)
And,
$3-\frac{x}{3}\le -5$...(2)
Step 2
Case 1 −
$3-\frac{x}{3}\ge 5$
$\frac{9-x}{3}\ge 5$
$9-x\ge 3×5$
$9-x\ge 15$
$9-15\ge x$
$x\le -6$...(3)
Case 2 −
$3-\frac{x}{3}\le -5$
$\frac{9-x}{3}\le -5$
$9-x\le -15$
$9+15\le x$
$x\ge 24$...(4)
From equation (3) and (4),
$x=\left(-\mathrm{\infty },-6\right]\cup \left[24,\mathrm{\infty }\right)$

Linda Birchfield
Step 1
Given: $4+|3-\frac{x}{3}|\ge 9$
$⇒|3-\frac{x}{3}|\ge 5$
$⇒|9-x|\ge 15$
Step 2
$⇒|x-9|\ge 15$

$⇒\left(-\mathrm{\infty },-6\right]\cup \left[24,\mathrm{\infty }\right)$.