Solve for n 4(8-4n)=2(n-40)+4

$4 (8 - 4 n) = 2 (n - 40) + 4$

Step 1

Simplify $4 (8 - 4 n)$ .

Rewrite.

$0 + 0 + 4 (8 - 4 n) = 2 (n - 40) + 4$

Simplify by adding zeros.

$4 (8 - 4 n) = 2 (n - 40) + 4$

Apply the distributive property.

$4 \cdot 8 + 4 (- 4 n) = 2 (n - 40) + 4$

Multiply.

Multiply $4$ by $8$ .

$32 + 4 (- 4 n) = 2 (n - 40) + 4$

Multiply $- 4$ by $4$ .

$32 - 16 n = 2 (n - 40) + 4$

Step 2

Simplify $2 (n - 40) + 4$ .

Simplify each term.

Apply the distributive property.

$32 - 16 n = 2 n + 2 \cdot - 40 + 4$

Multiply $2$ by $- 40$ .

$32 - 16 n = 2 n - 80 + 4$

Add $- 80$ and $4$ .

$32 - 16 n = 2 n - 76$

Step 3

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

Subtract $2 n$ from both sides of the equation.

$32 - 16 n - 2 n = - 76$

Subtract $2 n$ from $- 16 n$ .

$32 - 18 n = - 76$

Step 4

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

Subtract $32$ from both sides of the equation.

$- 18 n = - 76 - 32$

Subtract $32$ from $- 76$ .

$- 18 n = - 108$

Step 5

Divide each term in $- 18 n = - 108$ by $- 18$ and simplify.

Divide each term in $- 18 n = - 108$ by $- 18$ .

$\frac{- 18 n}{- 18} = \frac{- 108}{- 18}$

Simplify the left side.

Cancel the common factor of $- 18$ .

Cancel the common factor.

$\frac{- 18 n}{- 18} = \frac{- 108}{- 18}$

Divide $n$ by $1$ .

$n = \frac{- 108}{- 18}$

Simplify the right side.

Divide $- 108$ by $- 18$ .

$n = 6$

Solve for n 4(8-4n)=2(n-40)+4

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