Solve for r 2(3r+4)-3(r+1)=11

$2 (3 r + 4) - 3 (r + 1) = 11$

Simplify $2 (3 r + 4) - 3 (r + 1)$ .

Simplify each term.

Apply the distributive property.

$2 (3 r) + 2 \cdot 4 - 3 (r + 1) = 11$

Multiply $3$ by $2$ .

$6 r + 2 \cdot 4 - 3 (r + 1) = 11$

Multiply $2$ by $4$ .

$6 r + 8 - 3 (r + 1) = 11$

Apply the distributive property.

$6 r + 8 - 3 r - 3 \cdot 1 = 11$

Multiply $- 3$ by $1$ .

$6 r + 8 - 3 r - 3 = 11$

Simplify by adding terms.

Subtract $3 r$ from $6 r$ .

$3 r + 8 - 3 = 11$

Subtract $3$ from $8$ .

$3 r + 5 = 11$

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

Subtract $5$ from both sides of the equation.

$3 r = 11 - 5$

Subtract $5$ from $11$ .

$3 r = 6$

Divide each term in $3 r = 6$ by $3$ and simplify.

Divide each term in $3 r = 6$ by $3$ .

$\frac{3 r}{3} = \frac{6}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 r}{3} = \frac{6}{3}$

Divide $r$ by $1$ .

$r = \frac{6}{3}$

Simplify the right side.

Divide $6$ by $3$ .

$r = 2$