Solve {l}{x+y=16}{x^2+y^2=64}

$$\left\{ \begin{array} { l } { x + y = 16 } \\ { x ^ { 2 } + y ^ { 2 } = 64 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=8+4\sqrt{2}i\approx 8+5.656854249i\text{, }y=-4\sqrt{2}i+8\approx 8-5.656854249i$$

$$x=-4\sqrt{2}i+8\approx 8-5.656854249i\text{, }y=8+4\sqrt{2}i\approx 8+5.656854249i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=16 for x by isolating x on the left hand side of the equal sign.

$$x+y=16$$

Subtract y from both sides of the equation.

$$x=-y+16$$

Substitute -y+16 for x in the other equation, y^{2}+x^{2}=64.

$$y^{2}+\left(-y+16\right)^{2}=64$$

Square -y+16.

$$y^{2}+y^{2}-32y+256=64$$

Add y^{2} to y^{2}.

$$2y^{2}-32y+256=64$$

Subtract 64 from both sides of the equation.

$$2y^{2}-32y+192=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 16\left(-1\right)\times 2 for b, and 192 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 2\times 192}}{2\times 2}$$

Square 1\times 16\left(-1\right)\times 2.

$$y=\frac{-\left(-32\right)±\sqrt{1024-4\times 2\times 192}}{2\times 2}$$

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

$$y=\frac{-\left(-32\right)±\sqrt{1024-8\times 192}}{2\times 2}$$

Multiply -8 times 192.

$$y=\frac{-\left(-32\right)±\sqrt{1024-1536}}{2\times 2}$$

Add 1024 to -1536.

$$y=\frac{-\left(-32\right)±\sqrt{-512}}{2\times 2}$$

Take the square root of -512.

$$y=\frac{-\left(-32\right)±16\sqrt{2}i}{2\times 2}$$

The opposite of 1\times 16\left(-1\right)\times 2 is 32.

$$y=\frac{32±16\sqrt{2}i}{2\times 2}$$

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

$$y=\frac{32±16\sqrt{2}i}{4}$$

Now solve the equation y=\frac{32±16\sqrt{2}i}{4} when ± is plus. Add 32 to 16i\sqrt{2}.

$$y=\frac{32+2^{\frac{9}{2}}i}{4}$$

Divide 32+i\times 2^{\frac{9}{2}} by 4.

$$y=8+2^{\frac{5}{2}}i$$

Now solve the equation y=\frac{32±16\sqrt{2}i}{4} when ± is minus. Subtract 16i\sqrt{2} from 32.

$$y=\frac{-2^{\frac{9}{2}}i+32}{4}$$

Divide 32-i\times 2^{\frac{9}{2}} by 4.

$$y=-2^{\frac{5}{2}}i+8$$

There are two solutions for y: 8+i\times 2^{\frac{5}{2}} and 8-i\times 2^{\frac{5}{2}}. Substitute 8+i\times 2^{\frac{5}{2}} for y in the equation x=-y+16 to find the corresponding solution for x that satisfies both equations.

$$x=-\left(8+2^{\frac{5}{2}}i\right)+16$$

Now substitute 8-i\times 2^{\frac{5}{2}} for y in the equation x=-y+16 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-2^{\frac{5}{2}}i+8\right)+16$$

The system is now solved.

$$x=-\left(8+2^{\frac{5}{2}}i\right)+16,y=8+2^{\frac{5}{2}}i\text{ or }x=-\left(-2^{\frac{5}{2}}i+8\right)+16,y=-2^{\frac{5}{2}}i+8$$