Solve {l}{x^2+y^2+3=9}{x+y=10}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } + 3 = 9 } \\ { x + y = 10 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=5+\sqrt{22}i\approx 5+4.69041576i\text{, }y=-\sqrt{22}i+5\approx 5-4.69041576i$$

$$x=-\sqrt{22}i+5\approx 5-4.69041576i\text{, }y=5+\sqrt{22}i\approx 5+4.69041576i$$

Steps Using Substitution and Quadratic Formula

Solve x+y=10 for x by isolating x on the left hand side of the equal sign.

$$x+y=10$$

Subtract y from both sides of the equation.

$$x=-y+10$$

Substitute -y+10 for x in the other equation, y^{2}+x^{2}+3=9.

$$y^{2}+\left(-y+10\right)^{2}+3=9$$

Square -y+10.

$$y^{2}+y^{2}-20y+100+3=9$$

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

$$2y^{2}-20y+100+3=9$$

Add 1\times 10^{2} to 3.

$$2y^{2}-20y+103=9$$

Subtract 9 from both sides of the equation.

$$2y^{2}-20y+94=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 10\left(-1\right)\times 2 for b, and 94 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 2\times 94}}{2\times 2}$$

Square 1\times 10\left(-1\right)\times 2.

$$y=\frac{-\left(-20\right)±\sqrt{400-4\times 2\times 94}}{2\times 2}$$

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

$$y=\frac{-\left(-20\right)±\sqrt{400-8\times 94}}{2\times 2}$$

Multiply -8 times 94.

$$y=\frac{-\left(-20\right)±\sqrt{400-752}}{2\times 2}$$

Add 400 to -752.

$$y=\frac{-\left(-20\right)±\sqrt{-352}}{2\times 2}$$

Take the square root of -352.

$$y=\frac{-\left(-20\right)±4\sqrt{22}i}{2\times 2}$$

The opposite of 1\times 10\left(-1\right)\times 2 is 20.

$$y=\frac{20±4\sqrt{22}i}{2\times 2}$$

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

$$y=\frac{20±4\sqrt{22}i}{4}$$

Now solve the equation y=\frac{20±4\sqrt{22}i}{4} when ± is plus. Add 20 to 4i\sqrt{22}.

$$y=\frac{20+4\sqrt{22}i}{4}$$

Divide 20+4i\sqrt{22} by 4.

$$y=5+\sqrt{22}i$$

Now solve the equation y=\frac{20±4\sqrt{22}i}{4} when ± is minus. Subtract 4i\sqrt{22} from 20.

$$y=\frac{-4\sqrt{22}i+20}{4}$$

Divide 20-4i\sqrt{22} by 4.

$$y=-\sqrt{22}i+5$$

There are two solutions for y: 5+i\sqrt{22} and 5-i\sqrt{22}. Substitute 5+i\sqrt{22} for y in the equation x=-y+10 to find the corresponding solution for x that satisfies both equations.

$$x=-\left(5+\sqrt{22}i\right)+10$$

Now substitute 5-i\sqrt{22} for y in the equation x=-y+10 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-\sqrt{22}i+5\right)+10$$

The system is now solved.

$$x=-\left(5+\sqrt{22}i\right)+10,y=5+\sqrt{22}i\text{ or }x=-\left(-\sqrt{22}i+5\right)+10,y=-\sqrt{22}i+5$$