Solve t_{1}=D/c-v+D/c+v=

$$t _ { 1 } = \frac { D } { c - v } + \frac { D } { c + v } =$$

Solve for D

$$\left\{\begin{matrix}D=\frac{t_{1}\left(c^{2}-v^{2}\right)}{2c}\text{, }&c\neq 0\text{ and }|c|\neq |v|\\D\in \mathrm{R}\text{, }&t_{1}=0\text{ and }c=0\text{ and }v\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(-v+c\right)\left(-v-c\right), the least common multiple of c-v,c+v.

$$t_{1}\left(-v+c\right)\left(-v-c\right)=\left(-v-c\right)D+\left(-c+v\right)D$$

Use the distributive property to multiply t_{1} by -v+c.

$$\left(-t_{1}v+t_{1}c\right)\left(-v-c\right)=\left(-v-c\right)D+\left(-c+v\right)D$$

Use the distributive property to multiply -t_{1}v+t_{1}c by -v-c and combine like terms.

$$v^{2}t_{1}-t_{1}c^{2}=\left(-v-c\right)D+\left(-c+v\right)D$$

Use the distributive property to multiply -v-c by D.

$$v^{2}t_{1}-t_{1}c^{2}=-vD-cD+\left(-c+v\right)D$$

Use the distributive property to multiply -c+v by D.

$$v^{2}t_{1}-t_{1}c^{2}=-vD-cD-cD+vD$$

Combine -cD and -cD to get -2cD.

$$v^{2}t_{1}-t_{1}c^{2}=-vD-2cD+vD$$

Combine -vD and vD to get 0.

$$v^{2}t_{1}-t_{1}c^{2}=-2cD$$

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

$$-2cD=v^{2}t_{1}-t_{1}c^{2}$$

The equation is in standard form.

$$\left(-2c\right)D=t_{1}v^{2}-t_{1}c^{2}$$

Divide both sides by -2c.

$$\frac{\left(-2c\right)D}{-2c}=\frac{t_{1}\left(v-c\right)\left(v+c\right)}{-2c}$$

Dividing by -2c undoes the multiplication by -2c.

$$D=\frac{t_{1}\left(v-c\right)\left(v+c\right)}{-2c}$$

Divide t_{1}\left(v-c\right)\left(v+c\right) by -2c.

$$D=-\frac{t_{1}\left(v-c\right)\left(v+c\right)}{2c}$$

Solve for c

$$\left\{\begin{matrix}c=\frac{\sqrt{\left(t_{1}v\right)^{2}+D^{2}}+D}{t_{1}}\text{; }c=\frac{-\sqrt{\left(t_{1}v\right)^{2}+D^{2}}+D}{t_{1}}\text{, }&v\neq 0\text{ and }D\neq 0\text{ and }t_{1}\neq 0\\c=0\text{, }&v\neq 0\text{ and }t_{1}=0\text{ and }D\neq 0\\c\in \mathrm{R}\setminus v,-v\text{, }&t_{1}=0\text{ and }D=0\end{matrix}\right.$$