Solve 1/(50-m)^2dm=0.002dt

$$\frac { 1 } { ( 50 - m ) ^ { 2 } } d m = 0.002 d t$$

Solve for d

$$\left\{\begin{matrix}\\d=0\text{, }&\text{unconditionally}\\d\in \mathrm{R}\text{, }&t=\frac{500m}{\left(m-50\right)^{2}}\text{ and }m\neq 50\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(m-50\right)^{2}.

$$1dm=0.002dt\left(m-50\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(m-50\right)^{2}.

$$1dm=0.002dt\left(m^{2}-100m+2500\right)$$

Use the distributive property to multiply 0.002dt by m^{2}-100m+2500.

$$1dm=0.002dtm^{2}-0.2dtm+5dt$$

Subtract 0.002dtm^{2} from both sides.

$$1dm-0.002dtm^{2}=-0.2dtm+5dt$$

Add 0.2dtm to both sides.

$$1dm-0.002dtm^{2}+0.2dtm=5dt$$

Subtract 5dt from both sides.

$$1dm-0.002dtm^{2}+0.2dtm-5dt=0$$

Reorder the terms.

$$dm-0.002dtm^{2}+0.2dmt-5dt=0$$

Combine all terms containing d.

$$\left(m-0.002tm^{2}+0.2mt-5t\right)d=0$$

The equation is in standard form.

$$\left(-\frac{tm^{2}}{500}+\frac{mt}{5}+m-5t\right)d=0$$

Divide 0 by m-0.002tm^{2}+0.2mt-5t.

$$d=0$$

Solve for m

$$\left\{\begin{matrix}\\m=0\text{; }m\neq 50\text{, }&\text{unconditionally}\\m=\frac{50\left(t+\sqrt{5\left(2t+5\right)}+5\right)}{t}\text{; }m=\frac{50\left(t-\sqrt{5\left(2t+5\right)}+5\right)}{t}\text{, }&t\neq 0\text{ and }t\geq -\frac{5}{2}\end{matrix}\right.$$