Solve {l}{x^2+y^2=9}{x+y=3+3}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 9 } \\ { x + y = 3 + 3 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{3\sqrt{2}i}{2}+3\approx 3+2.121320344i\text{, }y=-\frac{3\sqrt{2}i}{2}+3\approx 3-2.121320344i$$

$$x=-\frac{3\sqrt{2}i}{2}+3\approx 3-2.121320344i\text{, }y=\frac{3\sqrt{2}i}{2}+3\approx 3+2.121320344i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=6 for x by isolating x on the left hand side of the equal sign.

$$x+y=6$$

Subtract y from both sides of the equation.

$$x=-y+6$$

Substitute -y+6 for x in the other equation, y^{2}+x^{2}=9.

$$y^{2}+\left(-y+6\right)^{2}=9$$

Square -y+6.

$$y^{2}+y^{2}-12y+36=9$$

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

$$2y^{2}-12y+36=9$$

Subtract 9 from both sides of the equation.

$$2y^{2}-12y+27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 6\left(-1\right)\times 2 for b, and 27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 2\times 27}}{2\times 2}$$

Square 1\times 6\left(-1\right)\times 2.

$$y=\frac{-\left(-12\right)±\sqrt{144-4\times 2\times 27}}{2\times 2}$$

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

$$y=\frac{-\left(-12\right)±\sqrt{144-8\times 27}}{2\times 2}$$

Multiply -8 times 27.

$$y=\frac{-\left(-12\right)±\sqrt{144-216}}{2\times 2}$$

Add 144 to -216.

$$y=\frac{-\left(-12\right)±\sqrt{-72}}{2\times 2}$$

Take the square root of -72.

$$y=\frac{-\left(-12\right)±6\sqrt{2}i}{2\times 2}$$

The opposite of 1\times 6\left(-1\right)\times 2 is 12.

$$y=\frac{12±6\sqrt{2}i}{2\times 2}$$

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

$$y=\frac{12±6\sqrt{2}i}{4}$$

Now solve the equation y=\frac{12±6\sqrt{2}i}{4} when ± is plus. Add 12 to 6i\sqrt{2}.

$$y=\frac{12+6\sqrt{2}i}{4}$$

Divide 12+6i\sqrt{2} by 4.

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

Now solve the equation y=\frac{12±6\sqrt{2}i}{4} when ± is minus. Subtract 6i\sqrt{2} from 12.

$$y=\frac{-6\sqrt{2}i+12}{4}$$

Divide 12-6i\sqrt{2} by 4.

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

There are two solutions for y: 3+\frac{3i\sqrt{2}}{2} and 3-\frac{3i\sqrt{2}}{2}. Substitute 3+\frac{3i\sqrt{2}}{2} for y in the equation x=-y+6 to find the corresponding solution for x that satisfies both equations.

$$x=-\left(\frac{3\sqrt{2}i}{2}+3\right)+6$$

Now substitute 3-\frac{3i\sqrt{2}}{2} for y in the equation x=-y+6 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-\frac{3\sqrt{2}i}{2}+3\right)+6$$

The system is now solved.

$$x=-\left(\frac{3\sqrt{2}i}{2}+3\right)+6,y=\frac{3\sqrt{2}i}{2}+3\text{ or }x=-\left(-\frac{3\sqrt{2}i}{2}+3\right)+6,y=-\frac{3\sqrt{2}i}{2}+3$$