Solve the Inequality for r 3r+2(4r+2)<=2(6r+1)

$3 r + 2 (4 r + 2) \leq 2 (6 r + 1)$

Simplify $3 r + 2 (4 r + 2)$ .

Simplify each term.

Apply the distributive property.

$3 r + 2 (4 r) + 2 \cdot 2 \leq 2 (6 r + 1)$

Multiply $4$ by $2$ .

$3 r + 8 r + 2 \cdot 2 \leq 2 (6 r + 1)$

Multiply $2$ by $2$ .

$3 r + 8 r + 4 \leq 2 (6 r + 1)$

Add $3 r$ and $8 r$ .

$11 r + 4 \leq 2 (6 r + 1)$

Simplify $2 (6 r + 1)$ .

Apply the distributive property.

$11 r + 4 \leq 2 (6 r) + 2 \cdot 1$

Multiply.

Multiply $6$ by $2$ .

$11 r + 4 \leq 12 r + 2 \cdot 1$

Multiply $2$ by $1$ .

$11 r + 4 \leq 12 r + 2$

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

Subtract $12 r$ from both sides of the inequality.

$11 r + 4 - 12 r \leq 2$

Subtract $12 r$ from $11 r$ .

$- r + 4 \leq 2$

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

Subtract $4$ from both sides of the inequality.

$- r \leq 2 - 4$

Subtract $4$ from $2$ .

$- r \leq - 2$

Divide each term in $- r \leq - 2$ by $- 1$ and simplify.

Divide each term in $- r \leq - 2$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- r}{- 1} \geq \frac{- 2}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{r}{1} \geq \frac{- 2}{- 1}$

Divide $r$ by $1$ .

$r \geq \frac{- 2}{- 1}$

Simplify the right side.

Divide $- 2$ by $- 1$ .

$r \geq 2$

The result can be shown in multiple forms.

Inequality Form:

$r \geq 2$

Interval Notation:

$[2, \infty)$

Solve the Inequality for r 3r+2(4r+2)<=2(6r+1)

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