Solve for r -3(1+6r)=14-r

$- 3 (1 + 6 r) = 14 - r$

Step 1

Simplify $- 3 (1 + 6 r)$ .

Rewrite.

$0 + 0 - 3 (1 + 6 r) = 14 - r$

Simplify by adding zeros.

$- 3 (1 + 6 r) = 14 - r$

Apply the distributive property.

$- 3 \cdot 1 - 3 (6 r) = 14 - r$

Multiply.

Multiply $- 3$ by $1$ .

$- 3 - 3 (6 r) = 14 - r$

Multiply $6$ by $- 3$ .

$- 3 - 18 r = 14 - r$

Step 2

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

Add $r$ to both sides of the equation.

$- 3 - 18 r + r = 14$

Add $- 18 r$ and $r$ .

$- 3 - 17 r = 14$

Step 3

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

Add $3$ to both sides of the equation.

$- 17 r = 14 + 3$

Add $14$ and $3$ .

$- 17 r = 17$

Step 4

Divide each term in $- 17 r = 17$ by $- 17$ and simplify.

Divide each term in $- 17 r = 17$ by $- 17$ .

$\frac{- 17 r}{- 17} = \frac{17}{- 17}$

Simplify the left side.

Cancel the common factor of $- 17$ .

Cancel the common factor.

$\frac{- 17 r}{- 17} = \frac{17}{- 17}$

Divide $r$ by $1$ .

$r = \frac{17}{- 17}$

Simplify the right side.

Divide $17$ by $- 17$ .

$r = - 1$