Solve {l}{x+y=11}{x^2+y^2=41}

$$\left. \begin{array} { l } { x + y = 11 } \\ { x ^ { 2 } + y ^ { 2 } = 41 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=\frac{11+\sqrt{39}i}{2}\approx 5.5+3.122498999i\text{, }y=\frac{-\sqrt{39}i+11}{2}\approx 5.5-3.122498999i$$

$$x=\frac{-\sqrt{39}i+11}{2}\approx 5.5-3.122498999i\text{, }y=\frac{11+\sqrt{39}i}{2}\approx 5.5+3.122498999i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=11 for x by isolating x on the left hand side of the equal sign.

$$x+y=11$$

Subtract y from both sides of the equation.

$$x=-y+11$$

Substitute -y+11 for x in the other equation, y^{2}+x^{2}=41.

$$y^{2}+\left(-y+11\right)^{2}=41$$

Square -y+11.

$$y^{2}+y^{2}-22y+121=41$$

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

$$2y^{2}-22y+121=41$$

Subtract 41 from both sides of the equation.

$$2y^{2}-22y+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 11\left(-1\right)\times 2 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 2\times 80}}{2\times 2}$$

Square 1\times 11\left(-1\right)\times 2.

$$y=\frac{-\left(-22\right)±\sqrt{484-4\times 2\times 80}}{2\times 2}$$

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

$$y=\frac{-\left(-22\right)±\sqrt{484-8\times 80}}{2\times 2}$$

Multiply -8 times 80.

$$y=\frac{-\left(-22\right)±\sqrt{484-640}}{2\times 2}$$

Add 484 to -640.

$$y=\frac{-\left(-22\right)±\sqrt{-156}}{2\times 2}$$

Take the square root of -156.

$$y=\frac{-\left(-22\right)±2\sqrt{39}i}{2\times 2}$$

The opposite of 1\times 11\left(-1\right)\times 2 is 22.

$$y=\frac{22±2\sqrt{39}i}{2\times 2}$$

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

$$y=\frac{22±2\sqrt{39}i}{4}$$

Now solve the equation y=\frac{22±2\sqrt{39}i}{4} when ± is plus. Add 22 to 2i\sqrt{39}.

$$y=\frac{22+2\sqrt{39}i}{4}$$

Divide 22+2i\sqrt{39} by 4.

$$y=\frac{11+\sqrt{39}i}{2}$$

Now solve the equation y=\frac{22±2\sqrt{39}i}{4} when ± is minus. Subtract 2i\sqrt{39} from 22.

$$y=\frac{-2\sqrt{39}i+22}{4}$$

Divide 22-2i\sqrt{39} by 4.

$$y=\frac{-\sqrt{39}i+11}{2}$$

There are two solutions for y: \frac{11+i\sqrt{39}}{2} and \frac{11-i\sqrt{39}}{2}. Substitute \frac{11+i\sqrt{39}}{2} for y in the equation x=-y+11 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{11+\sqrt{39}i}{2}+11$$

Now substitute \frac{11-i\sqrt{39}}{2} for y in the equation x=-y+11 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{-\sqrt{39}i+11}{2}+11$$

The system is now solved.

$$x=-\frac{11+\sqrt{39}i}{2}+11,y=\frac{11+\sqrt{39}i}{2}\text{ or }x=-\frac{-\sqrt{39}i+11}{2}+11,y=\frac{-\sqrt{39}i+11}{2}$$