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

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

Solve for x, y (complex solution)

$$x=2+\sqrt{2}i\approx 2+1.414213562i\text{, }y=-\sqrt{2}i+2\approx 2-1.414213562i$$

$$x=-\sqrt{2}i+2\approx 2-1.414213562i\text{, }y=2+\sqrt{2}i\approx 2+1.414213562i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=4 for x by isolating x on the left hand side of the equal sign.

$$x+y=4$$

Subtract y from both sides of the equation.

$$x=-y+4$$

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

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

Square -y+4.

$$y^{2}+y^{2}-8y+16=4$$

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

$$2y^{2}-8y+16=4$$

Subtract 4 from both sides of the equation.

$$2y^{2}-8y+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 4\left(-1\right)\times 2 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\times 12}}{2\times 2}$$

Square 1\times 4\left(-1\right)\times 2.

$$y=\frac{-\left(-8\right)±\sqrt{64-4\times 2\times 12}}{2\times 2}$$

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

$$y=\frac{-\left(-8\right)±\sqrt{64-8\times 12}}{2\times 2}$$

Multiply -8 times 12.

$$y=\frac{-\left(-8\right)±\sqrt{64-96}}{2\times 2}$$

Add 64 to -96.

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

Take the square root of -32.

$$y=\frac{-\left(-8\right)±4\sqrt{2}i}{2\times 2}$$

The opposite of 1\times 4\left(-1\right)\times 2 is 8.

$$y=\frac{8±4\sqrt{2}i}{2\times 2}$$

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

$$y=\frac{8±4\sqrt{2}i}{4}$$

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

$$y=\frac{8+2^{\frac{5}{2}}i}{4}$$

Divide 8+i\times 2^{\frac{5}{2}} by 4.

$$y=2+\sqrt{2}i$$

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

$$y=\frac{-2^{\frac{5}{2}}i+8}{4}$$

Divide 8-i\times 2^{\frac{5}{2}} by 4.

$$y=-\sqrt{2}i+2$$

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

$$x=-\left(2+\sqrt{2}i\right)+4$$

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

$$x=-\left(-\sqrt{2}i+2\right)+4$$

The system is now solved.

$$x=-\left(2+\sqrt{2}i\right)+4,y=2+\sqrt{2}i\text{ or }x=-\left(-\sqrt{2}i+2\right)+4,y=-\sqrt{2}i+2$$