Solve y=1/rx+r

$$y = \frac { 1 } { r } x + r$$

Solve for x

$$x=-r\left(r-y\right)$$

$$r\neq 0$$

Solution Steps

Multiply both sides of the equation by r.

$$yr=1x+rr$$

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

$$yr=1x+r^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$1x+r^{2}=yr$$

Subtract r^{2} from both sides.

$$1x=yr-r^{2}$$

Reorder the terms.

$$x=ry-r^{2}$$

Solve for r (complex solution)

$$\left\{\begin{matrix}r=\frac{\sqrt{y^{2}-4x}+y}{2}\text{, }&\left(arg(y)<\pi \text{ and }y\neq 0\right)\text{ or }x\neq 0\\r=\frac{-\sqrt{y^{2}-4x}+y}{2}\text{, }&\left(|\frac{arg(y^{2})}{2}-arg(y)|\geq \pi \text{ and }y\neq 0\right)\text{ or }x\neq 0\end{matrix}\right.$$

Solve for r

$$\left\{\begin{matrix}r=\frac{\sqrt{y^{2}-4x}+y}{2}\text{, }&\left(x\leq \frac{y^{2}}{4}\text{ and }y>0\text{ and }y\geq 2\sqrt{x}\text{ and }x\geq 0\right)\text{ or }x<0\text{ or }\left(|y|\geq 2\sqrt{x}\text{ and }x\leq \frac{y^{2}}{4}\text{ and }x>0\right)\\r=\frac{-\sqrt{y^{2}-4x}+y}{2}\text{, }&\left(x\leq \frac{y^{2}}{4}\text{ and }y<0\text{ and }y\leq -2\sqrt{x}\text{ and }x\geq 0\right)\text{ or }x<0\text{ or }\left(|y|\geq 2\sqrt{x}\text{ and }x\leq \frac{y^{2}}{4}\text{ and }x>0\right)\end{matrix}\right.$$