Solve x/c-3=cx+6

$$\frac { x } { c - 3 } = c x + 6$$

Solve for x

$$x=-\frac{6\left(3-c\right)}{1+3c-c^{2}}$$

$$c\neq \frac{\sqrt{13}+3}{2}\text{ and }c\neq \frac{3-\sqrt{13}}{2}\text{ and }c\neq 3$$

Steps for Solving Linear Equation

Multiply both sides of the equation by c-3.

$$x=cx\left(c-3\right)+\left(c-3\right)\times 6$$

Use the distributive property to multiply cx by c-3.

$$x=xc^{2}-3cx+\left(c-3\right)\times 6$$

Use the distributive property to multiply c-3 by 6.

$$x=xc^{2}-3cx+6c-18$$

Subtract xc^{2} from both sides.

$$x-xc^{2}=-3cx+6c-18$$

Add 3cx to both sides.

$$x-xc^{2}+3cx=6c-18$$

Reorder the terms.

$$-xc^{2}+3cx+x=6c-18$$

Combine all terms containing x.

$$\left(-c^{2}+3c+1\right)x=6c-18$$

The equation is in standard form.

$$\left(1+3c-c^{2}\right)x=6c-18$$

Divide both sides by -c^{2}+3c+1.

$$\frac{\left(1+3c-c^{2}\right)x}{1+3c-c^{2}}=\frac{6c-18}{1+3c-c^{2}}$$

Dividing by -c^{2}+3c+1 undoes the multiplication by -c^{2}+3c+1.

$$x=\frac{6c-18}{1+3c-c^{2}}$$

Divide -18+6c by -c^{2}+3c+1.

$$x=\frac{6\left(c-3\right)}{1+3c-c^{2}}$$

Solve for c

$$c=-\frac{\sqrt{13x^{2}+36x+36}-3x+6}{2x}$$

$$c=-\frac{-\sqrt{13x^{2}+36x+36}-3x+6}{2x}\text{, }x\neq 0$$