Solve for k 6(k+2)+1=13

$6 (k + 2) + 1 = 13$

Step 1

Simplify $6 (k + 2) + 1$ .

Simplify each term.

Apply the distributive property.

$6 k + 6 \cdot 2 + 1 = 13$

Multiply $6$ by $2$ .

$6 k + 12 + 1 = 13$

Add $12$ and $1$ .

$6 k + 13 = 13$

Step 2

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

Subtract $13$ from both sides of the equation.

$6 k = 13 - 13$

Subtract $13$ from $13$ .

$6 k = 0$

Step 3

Divide each term in $6 k = 0$ by $6$ and simplify.

Divide each term in $6 k = 0$ by $6$ .

$\frac{6 k}{6} = \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 k}{6} = \frac{0}{6}$

Divide $k$ by $1$ .

$k = \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$k = 0$