Solve a/b+b/c=6

$$\frac{ a }{ b } + \frac{ b }{ c } =6$$

Solve for a

$$a=-\frac{b\left(b-6c\right)}{c}$$

$$b\neq 0\text{ and }c\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by bc, the least common multiple of b,c.

$$ca+bb=6bc$$

Multiply b and b to get b^{2}.

$$ca+b^{2}=6bc$$

Subtract b^{2} from both sides.

$$ca=6bc-b^{2}$$

Divide both sides by c.

$$\frac{ca}{c}=\frac{b\left(6c-b\right)}{c}$$

Dividing by c undoes the multiplication by c.

$$a=\frac{b\left(6c-b\right)}{c}$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-\sqrt{c\left(9c-a\right)}+3c\text{, }&\left(arg(c)\geq \pi \text{ or }a\neq 0\right)\text{ and }c\neq 0\\b=\sqrt{c\left(9c-a\right)}+3c\text{, }&\left(arg(c)<\pi \text{ or }a\neq 0\right)\text{ and }c\neq 0\end{matrix}\right.$$

Solve for b

$$\left\{\begin{matrix}b=-\sqrt{c\left(9c-a\right)}+3c\text{, }&\left(a\geq 9c\text{ and }c<0\right)\text{ or }\left(c\neq 0\text{ and }a=9c\right)\text{ or }\left(a\neq 0\text{ and }a\leq 9c\text{ and }c>0\right)\\b=\sqrt{c\left(9c-a\right)}+3c\text{, }&\left(c>0\text{ and }a\leq 9c\right)\text{ or }\left(c\neq 0\text{ and }a=9c\right)\text{ or }\left(a\neq 0\text{ and }a\geq 9c\text{ and }c<0\right)\end{matrix}\right.$$