Solve {l}{n-m=2}{m^2+n^2=8}

$$\left\{ \begin{array} { l } { n - m = 2 } \\ { m ^ { 2 } + n ^ { 2 } = 8 } \end{array} \right.$$

Solve for n, m

$$n=1-\sqrt{3}\approx -0.732050808\text{, }m=-\left(\sqrt{3}+1\right)\approx -2.732050808$$

$$n=\sqrt{3}+1\approx 2.732050808\text{, }m=\sqrt{3}-1\approx 0.732050808$$

Steps Using Substitution and Quadratic Formula

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.

$$n-m=2,m^{2}+n^{2}=8$$

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

$$n-m=2$$

Subtract -m from both sides of the equation.

$$n=m+2$$

Substitute m+2 for n in the other equation, m^{2}+n^{2}=8.

$$m^{2}+\left(m+2\right)^{2}=8$$

Square m+2.

$$m^{2}+m^{2}+4m+4=8$$

Add m^{2} to m^{2}.

$$2m^{2}+4m+4=8$$

Subtract 8 from both sides of the equation.

$$2m^{2}+4m-4=0$$

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

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

Square 1\times 2\times 1\times 2.

$$m=\frac{-4±\sqrt{16-4\times 2\left(-4\right)}}{2\times 2}$$

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

$$m=\frac{-4±\sqrt{16-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$m=\frac{-4±\sqrt{16+32}}{2\times 2}$$

Add 16 to 32.

$$m=\frac{-4±\sqrt{48}}{2\times 2}$$

Take the square root of 48.

$$m=\frac{-4±4\sqrt{3}}{2\times 2}$$

Multiply 2 times 1+1\times 1^{2}.

$$m=\frac{-4±4\sqrt{3}}{4}$$

Now solve the equation m=\frac{-4±4\sqrt{3}}{4} when ± is plus. Add -4 to 4\sqrt{3}.

$$m=\frac{4\sqrt{3}-4}{4}$$

Divide -4+4\sqrt{3} by 4.

$$m=\sqrt{3}-1$$

Now solve the equation m=\frac{-4±4\sqrt{3}}{4} when ± is minus. Subtract 4\sqrt{3} from -4.

$$m=\frac{-4\sqrt{3}-4}{4}$$

Divide -4-4\sqrt{3} by 4.

$$m=-\sqrt{3}-1$$

There are two solutions for m: -1+\sqrt{3} and -1-\sqrt{3}. Substitute -1+\sqrt{3} for m in the equation n=m+2 to find the corresponding solution for n that satisfies both equations.

$$n=\sqrt{3}-1+2$$

Add 1\left(-1+\sqrt{3}\right) to 2.

$$n=\sqrt{3}+1$$

Now substitute -1-\sqrt{3} for m in the equation n=m+2 and solve to find the corresponding solution for n that satisfies both equations.

$$n=-\sqrt{3}-1+2$$

Add 1\left(-1-\sqrt{3}\right) to 2.

$$n=1-\sqrt{3}$$

The system is now solved.

$$n=\sqrt{3}+1,m=\sqrt{3}-1\text{ or }n=1-\sqrt{3},m=-\sqrt{3}-1$$