Solve {l}{x+y=3}{x^2+y^2=2}

$$\left. \begin{array} { l } { x + y = 3 } \\ { x ^ { 2 } + y ^ { 2 } = 2 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{3+\sqrt{5}i}{2}\approx 1.5+1.118033989i\text{, }y=\frac{-\sqrt{5}i+3}{2}\approx 1.5-1.118033989i$$

$$x=\frac{-\sqrt{5}i+3}{2}\approx 1.5-1.118033989i\text{, }y=\frac{3+\sqrt{5}i}{2}\approx 1.5+1.118033989i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=3 for x by isolating x on the left hand side of the equal sign.

$$x+y=3$$

Subtract y from both sides of the equation.

$$x=-y+3$$

Substitute -y+3 for x in the other equation, y^{2}+x^{2}=2.

$$y^{2}+\left(-y+3\right)^{2}=2$$

Square -y+3.

$$y^{2}+y^{2}-6y+9=2$$

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

$$2y^{2}-6y+9=2$$

Subtract 2 from both sides of the equation.

$$2y^{2}-6y+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 3\left(-1\right)\times 2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 2\times 7}}{2\times 2}$$

Square 1\times 3\left(-1\right)\times 2.

$$y=\frac{-\left(-6\right)±\sqrt{36-4\times 2\times 7}}{2\times 2}$$

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

$$y=\frac{-\left(-6\right)±\sqrt{36-8\times 7}}{2\times 2}$$

Multiply -8 times 7.

$$y=\frac{-\left(-6\right)±\sqrt{36-56}}{2\times 2}$$

Add 36 to -56.

$$y=\frac{-\left(-6\right)±\sqrt{-20}}{2\times 2}$$

Take the square root of -20.

$$y=\frac{-\left(-6\right)±2\sqrt{5}i}{2\times 2}$$

The opposite of 1\times 3\left(-1\right)\times 2 is 6.

$$y=\frac{6±2\sqrt{5}i}{2\times 2}$$

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

$$y=\frac{6±2\sqrt{5}i}{4}$$

Now solve the equation y=\frac{6±2\sqrt{5}i}{4} when ± is plus. Add 6 to 2i\sqrt{5}.

$$y=\frac{6+2\sqrt{5}i}{4}$$

Divide 6+2i\sqrt{5} by 4.

$$y=\frac{3+\sqrt{5}i}{2}$$

Now solve the equation y=\frac{6±2\sqrt{5}i}{4} when ± is minus. Subtract 2i\sqrt{5} from 6.

$$y=\frac{-2\sqrt{5}i+6}{4}$$

Divide 6-2i\sqrt{5} by 4.

$$y=\frac{-\sqrt{5}i+3}{2}$$

There are two solutions for y: \frac{3+i\sqrt{5}}{2} and \frac{3-i\sqrt{5}}{2}. Substitute \frac{3+i\sqrt{5}}{2} for y in the equation x=-y+3 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{3+\sqrt{5}i}{2}+3$$

Now substitute \frac{3-i\sqrt{5}}{2} for y in the equation x=-y+3 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{5}i+3}{2}+3$$

The system is now solved.

$$x=-\frac{3+\sqrt{5}i}{2}+3,y=\frac{3+\sqrt{5}i}{2}\text{ or }x=-\frac{-\sqrt{5}i+3}{2}+3,y=\frac{-\sqrt{5}i+3}{2}$$