Solve the Inequality for r 3r+2(12r+7)<=5r-8

$3 r + 2 (12 r + 7) \leq 5 r - 8$

Step 1

Simplify $3 r + 2 (12 r + 7)$ .

Simplify each term.

Apply the distributive property.

$3 r + 2 (12 r) + 2 \cdot 7 \leq 5 r - 8$

Multiply $12$ by $2$ .

$3 r + 24 r + 2 \cdot 7 \leq 5 r - 8$

Multiply $2$ by $7$ .

$3 r + 24 r + 14 \leq 5 r - 8$

Add $3 r$ and $24 r$ .

$27 r + 14 \leq 5 r - 8$

Step 2

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

Subtract $5 r$ from both sides of the inequality.

$27 r + 14 - 5 r \leq - 8$

Subtract $5 r$ from $27 r$ .

$22 r + 14 \leq - 8$

Step 3

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

Subtract $14$ from both sides of the inequality.

$22 r \leq - 8 - 14$

Subtract $14$ from $- 8$ .

$22 r \leq - 22$

Step 4

Divide each term in $22 r \leq - 22$ by $22$ and simplify.

Divide each term in $22 r \leq - 22$ by $22$ .

$\frac{22 r}{22} \leq \frac{- 22}{22}$

Simplify the left side.

Cancel the common factor of $22$ .

Cancel the common factor.

$\frac{22 r}{22} \leq \frac{- 22}{22}$

Divide $r$ by $1$ .

$r \leq \frac{- 22}{22}$

Simplify the right side.

Divide $- 22$ by $22$ .

$r \leq - 1$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$r \leq - 1$

Interval Notation:

$(- \infty, - 1]$