Solve for b
\left\{\begin{matrix}b=-\frac{a}{3-c}\text{, }&c\neq 3\\b\in \mathrm{R}\text{, }&a=0\text{ and }c=3\end{matrix}\right.
Solve for a
a=b\left(c-3\right)
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a+3b-bc=0
Subtract bc from both sides.
3b-bc=-a
Subtract a from both sides. Anything subtracted from zero gives its negation.
\left(3-c\right)b=-a
Combine all terms containing b.
\frac{\left(3-c\right)b}{3-c}=-\frac{a}{3-c}
Divide both sides by 3-c.
b=-\frac{a}{3-c}
Dividing by 3-c undoes the multiplication by 3-c.
a=bc-3b
Subtract 3b from both sides.
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