Solve for A
A=\frac{B_{3}C-87}{5}
Solve for B_3
\left\{\begin{matrix}B_{3}=\frac{5A+87}{C}\text{, }&C\neq 0\\B_{3}\in \mathrm{R}\text{, }&A=-\frac{87}{5}\text{ and }C=0\end{matrix}\right.
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5A=CB_{3}-87
Subtract 87 from both sides.
5A=B_{3}C-87
The equation is in standard form.
\frac{5A}{5}=\frac{B_{3}C-87}{5}
Divide both sides by 5.
A=\frac{B_{3}C-87}{5}
Dividing by 5 undoes the multiplication by 5.
CB_{3}=5A+87
Swap sides so that all variable terms are on the left hand side.
\frac{CB_{3}}{C}=\frac{5A+87}{C}
Divide both sides by C.
B_{3}=\frac{5A+87}{C}
Dividing by C undoes the multiplication by C.
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