Solve {l}{x^2+y^2=4}{x+y=8}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 4 } \\ { x + y = 8 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=4+\sqrt{14}i\approx 4+3.741657387i\text{, }y=-\sqrt{14}i+4\approx 4-3.741657387i$$

$$x=-\sqrt{14}i+4\approx 4-3.741657387i\text{, }y=4+\sqrt{14}i\approx 4+3.741657387i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=8 for x by isolating x on the left hand side of the equal sign.

$$x+y=8$$

Subtract y from both sides of the equation.

$$x=-y+8$$

Substitute -y+8 for x in the other equation, y^{2}+x^{2}=4.

$$y^{2}+\left(-y+8\right)^{2}=4$$

Square -y+8.

$$y^{2}+y^{2}-16y+64=4$$

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

$$2y^{2}-16y+64=4$$

Subtract 4 from both sides of the equation.

$$2y^{2}-16y+60=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 8\left(-1\right)\times 2 for b, and 60 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 2\times 60}}{2\times 2}$$

Square 1\times 8\left(-1\right)\times 2.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 2\times 60}}{2\times 2}$$

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

$$y=\frac{-\left(-16\right)±\sqrt{256-8\times 60}}{2\times 2}$$

Multiply -8 times 60.

$$y=\frac{-\left(-16\right)±\sqrt{256-480}}{2\times 2}$$

Add 256 to -480.

$$y=\frac{-\left(-16\right)±\sqrt{-224}}{2\times 2}$$

Take the square root of -224.

$$y=\frac{-\left(-16\right)±4\sqrt{14}i}{2\times 2}$$

The opposite of 1\times 8\left(-1\right)\times 2 is 16.

$$y=\frac{16±4\sqrt{14}i}{2\times 2}$$

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

$$y=\frac{16±4\sqrt{14}i}{4}$$

Now solve the equation y=\frac{16±4\sqrt{14}i}{4} when ± is plus. Add 16 to 4i\sqrt{14}.

$$y=\frac{16+4\sqrt{14}i}{4}$$

Divide 16+4i\sqrt{14} by 4.

$$y=4+\sqrt{14}i$$

Now solve the equation y=\frac{16±4\sqrt{14}i}{4} when ± is minus. Subtract 4i\sqrt{14} from 16.

$$y=\frac{-4\sqrt{14}i+16}{4}$$

Divide 16-4i\sqrt{14} by 4.

$$y=-\sqrt{14}i+4$$

There are two solutions for y: 4+i\sqrt{14} and 4-i\sqrt{14}. Substitute 4+i\sqrt{14} for y in the equation x=-y+8 to find the corresponding solution for x that satisfies both equations.

$$x=-\left(4+\sqrt{14}i\right)+8$$

Now substitute 4-i\sqrt{14} for y in the equation x=-y+8 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-\sqrt{14}i+4\right)+8$$

The system is now solved.

$$x=-\left(4+\sqrt{14}i\right)+8,y=4+\sqrt{14}i\text{ or }x=-\left(-\sqrt{14}i+4\right)+8,y=-\sqrt{14}i+4$$