Solve x/c-1=cx+3

$$\frac { x } { c - 1 } = c x + 3$$

Solve for x

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

$$c\neq \frac{\sqrt{5}+1}{2}\text{ and }c\neq \frac{1-\sqrt{5}}{2}\text{ and }c\neq 1$$

Steps for Solving Linear Equation

Multiply both sides of the equation by c-1.

$$x=cx\left(c-1\right)+\left(c-1\right)\times 3$$

Use the distributive property to multiply cx by c-1.

$$x=xc^{2}-cx+\left(c-1\right)\times 3$$

Use the distributive property to multiply c-1 by 3.

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

Subtract xc^{2} from both sides.

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

Add cx to both sides.

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

Reorder the terms.

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

Combine all terms containing x.

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

The equation is in standard form.

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

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

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

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

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

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

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

Solve for c

$$c=-\frac{\sqrt{5x^{2}+6x+9}-x+3}{2x}$$

$$c=-\frac{-\sqrt{5x^{2}+6x+9}-x+3}{2x}\text{, }x\neq 0$$