Solve m^2+n^2=16text{and}m+n+4=0

$$m ^ { 2 } + n ^ { 2 } = 16 \text { and } m + n + 4 = 0$$

Solve for m, n

$$m=0\text{, }n=-4$$

$$m=-4\text{, }n=0$$

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.

$$m+n+4=0,n^{2}+m^{2}=16$$

Solve m+n+4=0 for m by isolating m on the left hand side of the equal sign.

$$m+n+4=0$$

Subtract 4 from both sides of the equation.

$$m+n=-4$$

Subtract n from both sides of the equation.

$$m=-n-4$$

Substitute -n-4 for m in the other equation, n^{2}+m^{2}=16.

$$n^{2}+\left(-n-4\right)^{2}=16$$

Square -n-4.

$$n^{2}+n^{2}+8n+16=16$$

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

$$2n^{2}+8n+16=16$$

Subtract 16 from both sides of the equation.

$$2n^{2}+8n=0$$

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

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

Take the square root of 8^{2}.

$$n=\frac{-8±8}{2\times 2}$$

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

$$n=\frac{-8±8}{4}$$

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

$$n=\frac{0}{4}$$

Divide 0 by 4.

$$n=0$$

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

$$n=-\frac{16}{4}$$

Divide -16 by 4.

$$n=-4$$

There are two solutions for n: 0 and -4. Substitute 0 for n in the equation m=-n-4 to find the corresponding solution for m that satisfies both equations.

$$m=-4$$

Now substitute -4 for n in the equation m=-n-4 and solve to find the corresponding solution for m that satisfies both equations.

$$m=-\left(-4\right)-4$$

Multiply -1 times -4.

$$m=4-4$$

Add -4\left(-1\right) to -4.

$$m=0$$

The system is now solved.

$$m=-4,n=0\text{ or }m=0,n=-4$$