Question

$$ax=b(c+x)$$

Solve for a (complex solution)

$\left\{\begin{matrix}a=\frac{b\left(x+c\right)}{x}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&\left(b=0\text{ or }c=0\right)\text{ and }x=0\end{matrix}\right.$

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Solve for b (complex solution)

$\left\{\begin{matrix}b=\frac{ax}{x+c}\text{, }&x\neq -c\\b\in \mathrm{C}\text{, }&\left(a=0\text{ and }x=-c\right)\text{ or }\left(x=0\text{ and }c=0\right)\end{matrix}\right.$

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Solve for a

$\left\{\begin{matrix}a=\frac{b\left(x+c\right)}{x}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&\left(b=0\text{ or }c=0\right)\text{ and }x=0\end{matrix}\right.$

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Solve for b

$\left\{\begin{matrix}b=\frac{ax}{x+c}\text{, }&x\neq -c\\b\in \mathrm{R}\text{, }&\left(a=0\text{ and }x=-c\right)\text{ or }\left(x=0\text{ and }c=0\right)\end{matrix}\right.$

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