Solve {l}{n^2=2m^2}{2=2m+n}

$$\left\{ \begin{array} { l } { n ^ { 2 } = 2 m ^ { 2 } } \\ { 2 = 2 m + n } \end{array} \right.$$

Solve for n, m

$$n=2\sqrt{2}-2\approx 0.828427125\text{, }m=2-\sqrt{2}\approx 0.585786438$$

$$n=-2\sqrt{2}-2\approx -4.828427125\text{, }m=\sqrt{2}+2\approx 3.414213562$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 2m^{2} from both sides.

$$n^{2}-2m^{2}=0$$

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

$$2m+n=2$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$2m+n=2,n^{2}-2m^{2}=0$$

Solve 2m+n=2 for m by isolating m on the left hand side of the equal sign.

$$2m+n=2$$

Subtract n from both sides of the equation.

$$2m=-n+2$$

Divide both sides by 2.

$$m=-\frac{1}{2}n+1$$

Substitute -\frac{1}{2}n+1 for m in the other equation, n^{2}-2m^{2}=0.

$$n^{2}-2\left(-\frac{1}{2}n+1\right)^{2}=0$$

Square -\frac{1}{2}n+1.

$$n^{2}-2\left(\frac{1}{4}n^{2}-n+1\right)=0$$

Multiply -2 times \frac{1}{4}n^{2}-n+1.

$$n^{2}-\frac{1}{2}n^{2}+2n-2=0$$

Add n^{2} to -\frac{1}{2}n^{2}.

$$\frac{1}{2}n^{2}+2n-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1-2\left(-\frac{1}{2}\right)^{2} for a, -2\left(-\frac{1}{2}\right)\times 2 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-2±\sqrt{2^{2}-4\times \frac{1}{2}\left(-2\right)}}{2\times \frac{1}{2}}$$

Square -2\left(-\frac{1}{2}\right)\times 2.

$$n=\frac{-2±\sqrt{4-4\times \frac{1}{2}\left(-2\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times 1-2\left(-\frac{1}{2}\right)^{2}.

$$n=\frac{-2±\sqrt{4-2\left(-2\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -2.

$$n=\frac{-2±\sqrt{4+4}}{2\times \frac{1}{2}}$$

Add 4 to 4.

$$n=\frac{-2±\sqrt{8}}{2\times \frac{1}{2}}$$

Take the square root of 8.

$$n=\frac{-2±2\sqrt{2}}{2\times \frac{1}{2}}$$

Multiply 2 times 1-2\left(-\frac{1}{2}\right)^{2}.

$$n=\frac{-2±2\sqrt{2}}{1}$$

Now solve the equation n=\frac{-2±2\sqrt{2}}{1} when ± is plus. Add -2 to 2\sqrt{2}.

$$n=\frac{2\sqrt{2}-2}{1}$$

Divide -2+2\sqrt{2} by 1.

$$n=2\sqrt{2}-2$$

Now solve the equation n=\frac{-2±2\sqrt{2}}{1} when ± is minus. Subtract 2\sqrt{2} from -2.

$$n=\frac{-2\sqrt{2}-2}{1}$$

Divide -2-2\sqrt{2} by 1.

$$n=-2\sqrt{2}-2$$

There are two solutions for n: -2+2\sqrt{2} and -2-2\sqrt{2}. Substitute -2+2\sqrt{2} for n in the equation m=-\frac{1}{2}n+1 to find the corresponding solution for m that satisfies both equations.

$$m=-\frac{1}{2}\left(2\sqrt{2}-2\right)+1$$

Multiply -\frac{1}{2} times -2+2\sqrt{2}.

$$m=-\frac{2\sqrt{2}-2}{2}+1$$

Now substitute -2-2\sqrt{2} for n in the equation m=-\frac{1}{2}n+1 and solve to find the corresponding solution for m that satisfies both equations.

$$m=-\frac{1}{2}\left(-2\sqrt{2}-2\right)+1$$

Multiply -\frac{1}{2} times -2-2\sqrt{2}.

$$m=-\frac{-2\sqrt{2}-2}{2}+1$$

The system is now solved.

$$m=-\frac{2\sqrt{2}-2}{2}+1,n=2\sqrt{2}-2\text{ or }m=-\frac{-2\sqrt{2}-2}{2}+1,n=-2\sqrt{2}-2$$