Solve for F F/(5n^3)+h=g

$\frac{F}{5 n^{3}} + h = g$

Step 1

Subtract $h$ from both sides of the equation.

$\frac{F}{5 n^{3}} = g - h$

Step 2

Multiply both sides by $5 n^{3}$ .

$\frac{F}{5 n^{3}} (5 n^{3}) = (g - h) (5 n^{3})$

Step 3

Simplify.

Simplify the left side.

Simplify $\frac{F}{5 n^{3}} (5 n^{3})$ .

Rewrite using the commutative property of multiplication.

$5 \frac{F}{5 n^{3}} n^{3} = (g - h) (5 n^{3})$

Cancel the common factor of $5$ .

Factor $5$ out of $5 n^{3}$ .

$5 \frac{F}{5 (n^{3})} n^{3} = (g - h) (5 n^{3})$

Cancel the common factor.

$5 \frac{F}{5 n^{3}} n^{3} = (g - h) (5 n^{3})$

Rewrite the expression.

$\frac{F}{n^{3}} n^{3} = (g - h) (5 n^{3})$

Cancel the common factor of $n^{3}$ .

Cancel the common factor.

$\frac{F}{n^{3}} n^{3} = (g - h) (5 n^{3})$

Rewrite the expression.

$F = (g - h) (5 n^{3})$

Simplify the right side.

Simplify $(g - h) (5 n^{3})$ .

Apply the distributive property.

$F = g (5 n^{3}) - h (5 n^{3})$

Simplify the expression.

Rewrite using the commutative property of multiplication.

$F = 5 g n^{3} - h (5 n^{3})$

Rewrite using the commutative property of multiplication.

$F = 5 g n^{3} - 1 \cdot 5 h n^{3}$

Multiply $- 1$ by $5$ .

$F = 5 g n^{3} - 5 h n^{3}$