Solve for x -4(8-3x)>=6x-8

$- 4 (8 - 3 x) \geq 6 x - 8$

Step 1

Simplify $- 4 (8 - 3 x)$ .

Rewrite.

$0 + 0 - 4 (8 - 3 x) \geq 6 x - 8$

Simplify by adding zeros.

$- 4 (8 - 3 x) \geq 6 x - 8$

Apply the distributive property.

$- 4 \cdot 8 - 4 (- 3 x) \geq 6 x - 8$

Multiply.

Multiply $- 4$ by $8$ .

$- 32 - 4 (- 3 x) \geq 6 x - 8$

Multiply $- 3$ by $- 4$ .

$- 32 + 12 x \geq 6 x - 8$

Step 2

Move all terms containing $x$ to the left side of the inequality.

Subtract $6 x$ from both sides of the inequality.

$- 32 + 12 x - 6 x \geq - 8$

Subtract $6 x$ from $12 x$ .

$- 32 + 6 x \geq - 8$

Step 3

Move all terms not containing $x$ to the right side of the inequality.

Add $32$ to both sides of the inequality.

$6 x \geq - 8 + 32$

Add $- 8$ and $32$ .

$6 x \geq 24$

Step 4

Divide each term in $6 x \geq 24$ by $6$ and simplify.

Divide each term in $6 x \geq 24$ by $6$ .

$\frac{6 x}{6} \geq \frac{24}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} \geq \frac{24}{6}$

Divide $x$ by $1$ .

$x \geq \frac{24}{6}$

Simplify the right side.

Divide $24$ by $6$ .

$x \geq 4$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$x \geq 4$

Interval Notation:

$[4, \infty)$