Solve 1/t-4=t+x/r

$$\frac { 1 } { t } - 4 = \frac { t + x } { r }$$

Solve for r

$$\left\{\begin{matrix}r=-\frac{t\left(x+t\right)}{4t-1}\text{, }&t\neq -x\text{ and }t\neq 0\text{ and }t\neq \frac{1}{4}\\r\neq 0\text{, }&x=-\frac{1}{4}\text{ and }t=\frac{1}{4}\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable r cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by rt, the least common multiple of t,r.

$$r+rt\left(-4\right)=t\left(t+x\right)$$

Use the distributive property to multiply t by t+x.

$$r+rt\left(-4\right)=t^{2}+tx$$

Combine all terms containing r.

$$\left(1+t\left(-4\right)\right)r=t^{2}+tx$$

The equation is in standard form.

$$\left(1-4t\right)r=tx+t^{2}$$

Divide both sides by 1-4t.

$$\frac{\left(1-4t\right)r}{1-4t}=\frac{t\left(x+t\right)}{1-4t}$$

Dividing by 1-4t undoes the multiplication by 1-4t.

$$r=\frac{t\left(x+t\right)}{1-4t}$$

Variable r cannot be equal to 0.

$$r=\frac{t\left(x+t\right)}{1-4t}\text{, }r\neq 0$$

Solve for t (complex solution)

$$t=-\frac{\sqrt{x^{2}+8rx+16r^{2}+4r}}{2}-\frac{x}{2}-2r$$

$$t=\frac{\sqrt{x^{2}+8rx+16r^{2}+4r}}{2}-\frac{x}{2}-2r\text{, }r\neq 0$$

Solve for t

$$t=-\frac{\sqrt{x^{2}+8rx+16r^{2}+4r}}{2}-\frac{x}{2}-2r$$

$$t=\frac{\sqrt{x^{2}+8rx+16r^{2}+4r}}{2}-\frac{x}{2}-2r\text{, }r\neq 0\text{ and }\left(r>0\text{ or }x\leq -4r-2\sqrt{-r}\text{ or }x\geq -4r+2\sqrt{-r}\right)$$