Solve {l}{x^2/2+y^2=1}{x+y=2}

$$\left. \begin{array} { l } { \frac { x ^ { 2 } } { 2 } + y ^ { 2 } = 1 } \\ { x + y = 2 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{4+\sqrt{2}i}{3}\approx 1.333333333+0.471404521i\text{, }y=\frac{-\sqrt{2}i+2}{3}\approx 0.666666667-0.471404521i$$

$$x=\frac{-\sqrt{2}i+4}{3}\approx 1.333333333-0.471404521i\text{, }y=\frac{2+\sqrt{2}i}{3}\approx 0.666666667+0.471404521i$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 2.

$$x^{2}+2y^{2}=2$$

Solve x+y=2 for x by isolating x on the left hand side of the equal sign.

$$x+y=2$$

Subtract y from both sides of the equation.

$$x=-y+2$$

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

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

Square -y+2.

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

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

$$3y^{2}-4y+4=2$$

Subtract 2 from both sides of the equation.

$$3y^{2}-4y+2=0$$

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

$$y=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 3\times 2}}{2\times 3}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-4\times 3\times 2}}{2\times 3}$$

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

$$y=\frac{-\left(-4\right)±\sqrt{16-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$y=\frac{-\left(-4\right)±\sqrt{16-24}}{2\times 3}$$

Add 16 to -24.

$$y=\frac{-\left(-4\right)±\sqrt{-8}}{2\times 3}$$

Take the square root of -8.

$$y=\frac{-\left(-4\right)±2\sqrt{2}i}{2\times 3}$$

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

$$y=\frac{4±2\sqrt{2}i}{2\times 3}$$

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

$$y=\frac{4±2\sqrt{2}i}{6}$$

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

$$y=\frac{4+2\sqrt{2}i}{6}$$

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

$$y=\frac{2+\sqrt{2}i}{3}$$

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

$$y=\frac{-2\sqrt{2}i+4}{6}$$

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

$$y=\frac{-\sqrt{2}i+2}{3}$$

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

$$x=-\frac{2+\sqrt{2}i}{3}+2$$

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

$$x=-\frac{-\sqrt{2}i+2}{3}+2$$

The system is now solved.

$$x=-\frac{2+\sqrt{2}i}{3}+2,y=\frac{2+\sqrt{2}i}{3}\text{ or }x=-\frac{-\sqrt{2}i+2}{3}+2,y=\frac{-\sqrt{2}i+2}{3}$$