Solve the Inequality for r 3r-8>3(r-6)

$3 r - 8 > 3 (r - 6)$

Simplify $3 (r - 6)$ .

Rewrite.

$3 r - 8 > 0 + 0 + 3 (r - 6)$

Simplify by adding zeros.

$3 r - 8 > 3 (r - 6)$

Apply the distributive property.

$3 r - 8 > 3 r + 3 \cdot - 6$

Multiply $3$ by $- 6$ .

$3 r - 8 > 3 r - 18$

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

Subtract $3 r$ from both sides of the inequality.

$3 r - 8 - 3 r > - 18$

Combine the opposite terms in $3 r - 8 - 3 r$ .

Subtract $3 r$ from $3 r$ .

$0 - 8 > - 18$

Subtract $8$ from $0$ .

$- 8 > - 18$

Since $- 8 > - 18$ , the inequality will always be true.

Always true

The result can be shown in multiple forms.

Always true

Interval Notation:

$(- \infty, \infty)$