Solve for A 5gA+mA=B^3

$5 g A + mA = B^{3}$

Step 1

Subtract $mA$ from both sides of the equation.

$5 g A = B^{3} - mA$

Step 2

Divide each term in $5 g A = B^{3} - mA$ by $5 g$ and simplify.

Divide each term in $5 g A = B^{3} - mA$ by $5 g$ .

$\frac{5 g A}{5 g} = \frac{B^{3}}{5 g} + \frac{- mA}{5 g}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 g A}{5 g} = \frac{B^{3}}{5 g} + \frac{- mA}{5 g}$

Rewrite the expression.

$\frac{g A}{g} = \frac{B^{3}}{5 g} + \frac{- mA}{5 g}$

Cancel the common factor of $g$ .

Cancel the common factor.

$\frac{g A}{g} = \frac{B^{3}}{5 g} + \frac{- mA}{5 g}$

Divide $A$ by $1$ .

$A = \frac{B^{3}}{5 g} + \frac{- mA}{5 g}$

Simplify the right side.

Move the negative in front of the fraction.

$A = \frac{B^{3}}{5 g} - \frac{mA}{5 g}$

Step 3

Rewrite the expression.

$A = \frac{B^{3}}{5 g} - \frac{1}{5 g} mA$