Solve for r 4r-3=3(3r+4)

$4 r - 3 = 3 (3 r + 4)$

Simplify $3 (3 r + 4)$ .

Rewrite.

$4 r - 3 = 0 + 0 + 3 (3 r + 4)$

Simplify by adding zeros.

$4 r - 3 = 3 (3 r + 4)$

Apply the distributive property.

$4 r - 3 = 3 (3 r) + 3 \cdot 4$

Multiply.

Multiply $3$ by $3$ .

$4 r - 3 = 9 r + 3 \cdot 4$

Multiply $3$ by $4$ .

$4 r - 3 = 9 r + 12$

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

Subtract $9 r$ from both sides of the equation.

$4 r - 3 - 9 r = 12$

Subtract $9 r$ from $4 r$ .

$- 5 r - 3 = 12$

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

Add $3$ to both sides of the equation.

$- 5 r = 12 + 3$

Add $12$ and $3$ .

$- 5 r = 15$

Divide each term in $- 5 r = 15$ by $- 5$ and simplify.

Divide each term in $- 5 r = 15$ by $- 5$ .

$\frac{- 5 r}{- 5} = \frac{15}{- 5}$

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

$\frac{- 5 r}{- 5} = \frac{15}{- 5}$

Divide $r$ by $1$ .

$r = \frac{15}{- 5}$

Simplify the right side.

Divide $15$ by $- 5$ .

$r = - 3$