Solve for A BA+8=3mA

$B A + 8 = 3 mA$

Step 1

Subtract $8$ from both sides of the equation.

$B A = 3 mA - 8$

Step 2

Divide each term in $B A = 3 mA - 8$ by $B$ and simplify.

Divide each term in $B A = 3 mA - 8$ by $B$ .

$\frac{B A}{B} = \frac{3 mA}{B} + \frac{- 8}{B}$

Simplify the left side.

Cancel the common factor of $B$ .

Cancel the common factor.

$\frac{B A}{B} = \frac{3 mA}{B} + \frac{- 8}{B}$

Divide $A$ by $1$ .

$A = \frac{3 mA}{B} + \frac{- 8}{B}$

Simplify the right side.

Move the negative in front of the fraction.

$A = \frac{3 mA}{B} - \frac{8}{B}$

Step 3

Rewrite the expression.

$A = \frac{3}{B} mA - \frac{8}{B}$