Solve {l}{m^2+n^2=80}{m-n=4sqrt{2}}

$$\left. \begin{array} { l } { m ^ { 2 } + n ^ { 2 } = 80 } \\ { m - n = 4 \sqrt { 2 } } \end{array} \right.$$

Solve for m, n

$$m=-2\sqrt{2}\approx -2.828427125\text{, }n=-6\sqrt{2}\approx -8.485281374$$

$$m=6\sqrt{2}\approx 8.485281374\text{, }n=2\sqrt{2}\approx 2.828427125$$

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.

$$m-n=4\sqrt{2},n^{2}+m^{2}=80$$

Solve m-n=4\sqrt{2} for m by isolating m on the left hand side of the equal sign.

$$m-n=4\sqrt{2}$$

Subtract -n from both sides of the equation.

$$m=n+4\sqrt{2}$$

Substitute n+4\sqrt{2} for m in the other equation, n^{2}+m^{2}=80.

$$n^{2}+\left(n+4\sqrt{2}\right)^{2}=80$$

Square n+4\sqrt{2}.

$$n^{2}+n^{2}+8\sqrt{2}n+\left(4\sqrt{2}\right)^{2}=80$$

Add n^{2} to n^{2}.

$$2n^{2}+8\sqrt{2}n+\left(4\sqrt{2}\right)^{2}=80$$

Subtract 80 from both sides of the equation.

$$2n^{2}+8\sqrt{2}n+\left(4\sqrt{2}\right)^{2}-80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 1\times 2\times 4\sqrt{2} for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-8\sqrt{2}±\sqrt{\left(8\sqrt{2}\right)^{2}-4\times 2\left(-48\right)}}{2\times 2}$$

Square 1\times 1\times 2\times 4\sqrt{2}.

$$n=\frac{-8\sqrt{2}±\sqrt{128-4\times 2\left(-48\right)}}{2\times 2}$$

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

$$n=\frac{-8\sqrt{2}±\sqrt{128-8\left(-48\right)}}{2\times 2}$$

Multiply -8 times -48.

$$n=\frac{-8\sqrt{2}±\sqrt{128+384}}{2\times 2}$$

Add 128 to 384.

$$n=\frac{-8\sqrt{2}±\sqrt{512}}{2\times 2}$$

Take the square root of 512.

$$n=\frac{-8\sqrt{2}±16\sqrt{2}}{2\times 2}$$

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

$$n=\frac{-8\sqrt{2}±16\sqrt{2}}{4}$$

Now solve the equation n=\frac{-8\sqrt{2}±16\sqrt{2}}{4} when ± is plus. Add -8\sqrt{2} to 16\sqrt{2}.

$$n=\frac{8\sqrt{2}}{4}$$

Divide 8\sqrt{2} by 4.

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

Now solve the equation n=\frac{-8\sqrt{2}±16\sqrt{2}}{4} when ± is minus. Subtract 16\sqrt{2} from -8\sqrt{2}.

$$n=-\frac{24\sqrt{2}}{4}$$

Divide -24\sqrt{2} by 4.

$$n=-6\sqrt{2}$$

There are two solutions for n: 2\sqrt{2} and -6\sqrt{2}. Substitute 2\sqrt{2} for n in the equation m=n+4\sqrt{2} to find the corresponding solution for m that satisfies both equations.

$$m=2\sqrt{2}+4\sqrt{2}$$

Now substitute -6\sqrt{2} for n in the equation m=n+4\sqrt{2} and solve to find the corresponding solution for m that satisfies both equations.

$$m=-6\sqrt{2}+4\sqrt{2}$$

Add 1\left(-6\sqrt{2}\right) to 4\sqrt{2}.

$$m=4\sqrt{2}-6\sqrt{2}$$

The system is now solved.

$$m=2\sqrt{2}+4\sqrt{2},n=2\sqrt{2}\text{ or }m=4\sqrt{2}-6\sqrt{2},n=-6\sqrt{2}$$