Solve {l}{m^2+n^2=13}{m+n=3}

$$\left. \begin{array} { l } { m ^ { 2 } + n ^ { 2 } = 13 } \\ { m + n = 3 } \end{array} \right.$$

Solve for m, n

$$m=\frac{\sqrt{17}+3}{2}\approx 3.561552813\text{, }n=\frac{3-\sqrt{17}}{2}\approx -0.561552813$$

$$m=\frac{3-\sqrt{17}}{2}\approx -0.561552813\text{, }n=\frac{\sqrt{17}+3}{2}\approx 3.561552813$$

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.

$$m+n=3,n^{2}+m^{2}=13$$

Solve m+n=3 for m by isolating m on the left hand side of the equal sign.

$$m+n=3$$

Subtract n from both sides of the equation.

$$m=-n+3$$

Substitute -n+3 for m in the other equation, n^{2}+m^{2}=13.

$$n^{2}+\left(-n+3\right)^{2}=13$$

Square -n+3.

$$n^{2}+n^{2}-6n+9=13$$

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

$$2n^{2}-6n+9=13$$

Subtract 13 from both sides of the equation.

$$2n^{2}-6n-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 3\left(-1\right)\times 2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\left(-4\right)}}{2\times 2}$$

Square 1\times 3\left(-1\right)\times 2.

$$n=\frac{-\left(-6\right)±\sqrt{36-4\times 2\left(-4\right)}}{2\times 2}$$

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

$$n=\frac{-\left(-6\right)±\sqrt{36-8\left(-4\right)}}{2\times 2}$$

Multiply -8 times -4.

$$n=\frac{-\left(-6\right)±\sqrt{36+32}}{2\times 2}$$

Add 36 to 32.

$$n=\frac{-\left(-6\right)±\sqrt{68}}{2\times 2}$$

Take the square root of 68.

$$n=\frac{-\left(-6\right)±2\sqrt{17}}{2\times 2}$$

The opposite of 1\times 3\left(-1\right)\times 2 is 6.

$$n=\frac{6±2\sqrt{17}}{2\times 2}$$

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

$$n=\frac{6±2\sqrt{17}}{4}$$

Now solve the equation n=\frac{6±2\sqrt{17}}{4} when ± is plus. Add 6 to 2\sqrt{17}.

$$n=\frac{2\sqrt{17}+6}{4}$$

Divide 6+2\sqrt{17} by 4.

$$n=\frac{\sqrt{17}+3}{2}$$

Now solve the equation n=\frac{6±2\sqrt{17}}{4} when ± is minus. Subtract 2\sqrt{17} from 6.

$$n=\frac{6-2\sqrt{17}}{4}$$

Divide 6-2\sqrt{17} by 4.

$$n=\frac{3-\sqrt{17}}{2}$$

There are two solutions for n: \frac{3+\sqrt{17}}{2} and \frac{3-\sqrt{17}}{2}. Substitute \frac{3+\sqrt{17}}{2} for n in the equation m=-n+3 to find the corresponding solution for m that satisfies both equations.

$$m=-\frac{\sqrt{17}+3}{2}+3$$

Now substitute \frac{3-\sqrt{17}}{2} for n in the equation m=-n+3 and solve to find the corresponding solution for m that satisfies both equations.

$$m=-\frac{3-\sqrt{17}}{2}+3$$

The system is now solved.

$$m=-\frac{\sqrt{17}+3}{2}+3,n=\frac{\sqrt{17}+3}{2}\text{ or }m=-\frac{3-\sqrt{17}}{2}+3,n=\frac{3-\sqrt{17}}{2}$$