Solve for n 39=6(6n-3)-3(3n-1)

$39 = 6 (6 n - 3) - 3 (3 n - 1)$

Rewrite the equation as $6 (6 n - 3) - 3 (3 n - 1) = 39$ .

$6 (6 n - 3) - 3 (3 n - 1) = 39$

Simplify $6 (6 n - 3) - 3 (3 n - 1)$ .

Simplify each term.

Apply the distributive property.

$6 (6 n) + 6 \cdot - 3 - 3 (3 n - 1) = 39$

Multiply $6$ by $6$ .

$36 n + 6 \cdot - 3 - 3 (3 n - 1) = 39$

Multiply $6$ by $- 3$ .

$36 n - 18 - 3 (3 n - 1) = 39$

Apply the distributive property.

$36 n - 18 - 3 (3 n) - 3 \cdot - 1 = 39$

Multiply $3$ by $- 3$ .

$36 n - 18 - 9 n - 3 \cdot - 1 = 39$

Multiply $- 3$ by $- 1$ .

$36 n - 18 - 9 n + 3 = 39$

Simplify by adding terms.

Subtract $9 n$ from $36 n$ .

$27 n - 18 + 3 = 39$

Add $- 18$ and $3$ .

$27 n - 15 = 39$

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

Add $15$ to both sides of the equation.

$27 n = 39 + 15$

Add $39$ and $15$ .

$27 n = 54$

Divide each term in $27 n = 54$ by $27$ and simplify.

Divide each term in $27 n = 54$ by $27$ .

$\frac{27 n}{27} = \frac{54}{27}$

Simplify the left side.

Cancel the common factor of $27$ .

Cancel the common factor.

$\frac{27 n}{27} = \frac{54}{27}$

Divide $n$ by $1$ .

$n = \frac{54}{27}$

Simplify the right side.

Divide $54$ by $27$ .

$n = 2$