# How do you solve for A in B=5/7(A-9)?

May 22, 2017

See a solution process below:

#### Explanation:

First, multiply each side of the equation by $\frac{\textcolor{red}{7}}{\textcolor{b l u e}{5}}$ to eliminate the need for parenthesis while keeping the equation balanced:

$\frac{\textcolor{red}{7}}{\textcolor{b l u e}{5}} \times B = \frac{\textcolor{red}{7}}{\textcolor{b l u e}{5}} \times \frac{5}{7} \left(A - 9\right)$

$\frac{7}{5} B = \frac{\cancel{\textcolor{red}{7}}}{\cancel{\textcolor{b l u e}{5}}} \times \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{5}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{7}}}} \left(A - 9\right)$

$\frac{7}{5} B = A - 9$

Now, add $\textcolor{red}{9}$ to each side of the equation to solve for $A$ while keeping the equation balanced:

$\frac{7}{5} B + \textcolor{red}{9} = A - 9 + \textcolor{red}{9}$

$\frac{7}{5} B + 9 = A - 0$

$\frac{7}{5} B + 9 = A$

$A = \frac{7}{5} B + 9$