Solve {l}{x^2+y^2=34}{x+y=9}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 34 } \\ { x + y = 9 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{9+\sqrt{13}i}{2}\approx 4.5+1.802775638i\text{, }y=\frac{-\sqrt{13}i+9}{2}\approx 4.5-1.802775638i$$

$$x=\frac{-\sqrt{13}i+9}{2}\approx 4.5-1.802775638i\text{, }y=\frac{9+\sqrt{13}i}{2}\approx 4.5+1.802775638i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=9 for x by isolating x on the left hand side of the equal sign.

$$x+y=9$$

Subtract y from both sides of the equation.

$$x=-y+9$$

Substitute -y+9 for x in the other equation, y^{2}+x^{2}=34.

$$y^{2}+\left(-y+9\right)^{2}=34$$

Square -y+9.

$$y^{2}+y^{2}-18y+81=34$$

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

$$2y^{2}-18y+81=34$$

Subtract 34 from both sides of the equation.

$$2y^{2}-18y+47=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 9\left(-1\right)\times 2 for b, and 47 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 2\times 47}}{2\times 2}$$

Square 1\times 9\left(-1\right)\times 2.

$$y=\frac{-\left(-18\right)±\sqrt{324-4\times 2\times 47}}{2\times 2}$$

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

$$y=\frac{-\left(-18\right)±\sqrt{324-8\times 47}}{2\times 2}$$

Multiply -8 times 47.

$$y=\frac{-\left(-18\right)±\sqrt{324-376}}{2\times 2}$$

Add 324 to -376.

$$y=\frac{-\left(-18\right)±\sqrt{-52}}{2\times 2}$$

Take the square root of -52.

$$y=\frac{-\left(-18\right)±2\sqrt{13}i}{2\times 2}$$

The opposite of 1\times 9\left(-1\right)\times 2 is 18.

$$y=\frac{18±2\sqrt{13}i}{2\times 2}$$

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

$$y=\frac{18±2\sqrt{13}i}{4}$$

Now solve the equation y=\frac{18±2\sqrt{13}i}{4} when ± is plus. Add 18 to 2i\sqrt{13}.

$$y=\frac{18+2\sqrt{13}i}{4}$$

Divide 18+2i\sqrt{13} by 4.

$$y=\frac{9+\sqrt{13}i}{2}$$

Now solve the equation y=\frac{18±2\sqrt{13}i}{4} when ± is minus. Subtract 2i\sqrt{13} from 18.

$$y=\frac{-2\sqrt{13}i+18}{4}$$

Divide 18-2i\sqrt{13} by 4.

$$y=\frac{-\sqrt{13}i+9}{2}$$

There are two solutions for y: \frac{9+i\sqrt{13}}{2} and \frac{9-i\sqrt{13}}{2}. Substitute \frac{9+i\sqrt{13}}{2} for y in the equation x=-y+9 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{9+\sqrt{13}i}{2}+9$$

Now substitute \frac{9-i\sqrt{13}}{2} for y in the equation x=-y+9 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{13}i+9}{2}+9$$

The system is now solved.

$$x=-\frac{9+\sqrt{13}i}{2}+9,y=\frac{9+\sqrt{13}i}{2}\text{ or }x=-\frac{-\sqrt{13}i+9}{2}+9,y=\frac{-\sqrt{13}i+9}{2}$$