Solve (m^2+1/m^2=79+ms)

$$( m ^ { 2 } + \frac { 1 } { m ^ { 2 } } = 79 + m s )$$

Solve for s

$$s=m-\frac{79}{m}+\frac{1}{m^{3}}$$

$$m\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by m^{2}.

$$m^{2}m^{2}+1=m^{2}\times 79+msm^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$m^{4}+1=m^{2}\times 79+msm^{2}$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$m^{4}+1=m^{2}\times 79+m^{3}s$$

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

$$m^{2}\times 79+m^{3}s=m^{4}+1$$

Subtract m^{2}\times 79 from both sides.

$$m^{3}s=m^{4}+1-m^{2}\times 79$$

Multiply -1 and 79 to get -79.

$$m^{3}s=m^{4}+1-79m^{2}$$

The equation is in standard form.

$$m^{3}s=m^{4}-79m^{2}+1$$

Divide both sides by m^{3}.

$$\frac{m^{3}s}{m^{3}}=\frac{m^{4}-79m^{2}+1}{m^{3}}$$

Dividing by m^{3} undoes the multiplication by m^{3}.

$$s=\frac{m^{4}-79m^{2}+1}{m^{3}}$$

Divide -79m^{2}+m^{4}+1 by m^{3}.

$$s=m-\frac{79}{m}+\frac{1}{m^{3}}$$