Solve {l}{x^2+y^2=10}{x-y-2=10}

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

Solve for x, y (complex solution)

$$x=-\sqrt{31}i+6\approx 6-5.567764363i\text{, }y=-\sqrt{31}i-6\approx -6-5.567764363i$$

$$x=6+\sqrt{31}i\approx 6+5.567764363i\text{, }y=-6+\sqrt{31}i\approx -6+5.567764363i$$

Steps Using Substitution and Quadratic Formula

Solve x-y-2=10 for x by isolating x on the left hand side of the equal sign.

$$x-y-2=10$$

Add 2 to both sides of the equation.

$$x-y=12$$

Subtract -y from both sides of the equation.

$$x=y+12$$

Substitute y+12 for x in the other equation, y^{2}+x^{2}=10.

$$y^{2}+\left(y+12\right)^{2}=10$$

Square y+12.

$$y^{2}+y^{2}+24y+144=10$$

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

$$2y^{2}+24y+144=10$$

Subtract 10 from both sides of the equation.

$$2y^{2}+24y+134=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 12\times 1\times 2 for b, and 134 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-24±\sqrt{24^{2}-4\times 2\times 134}}{2\times 2}$$

Square 1\times 12\times 1\times 2.

$$y=\frac{-24±\sqrt{576-4\times 2\times 134}}{2\times 2}$$

Multiply -4 times 1+1\times 1^{2}.

$$y=\frac{-24±\sqrt{576-8\times 134}}{2\times 2}$$

Multiply -8 times 134.

$$y=\frac{-24±\sqrt{576-1072}}{2\times 2}$$

Add 576 to -1072.

$$y=\frac{-24±\sqrt{-496}}{2\times 2}$$

Take the square root of -496.

$$y=\frac{-24±4\sqrt{31}i}{2\times 2}$$

Multiply 2 times 1+1\times 1^{2}.

$$y=\frac{-24±4\sqrt{31}i}{4}$$

Now solve the equation y=\frac{-24±4\sqrt{31}i}{4} when ± is plus. Add -24 to 4i\sqrt{31}.

$$y=\frac{-24+4\sqrt{31}i}{4}$$

Divide -24+4i\sqrt{31} by 4.

$$y=-6+\sqrt{31}i$$

Now solve the equation y=\frac{-24±4\sqrt{31}i}{4} when ± is minus. Subtract 4i\sqrt{31} from -24.

$$y=\frac{-4\sqrt{31}i-24}{4}$$

Divide -24-4i\sqrt{31} by 4.

$$y=-\sqrt{31}i-6$$

There are two solutions for y: -6+i\sqrt{31} and -6-i\sqrt{31}. Substitute -6+i\sqrt{31} for y in the equation x=y+12 to find the corresponding solution for x that satisfies both equations.

$$x=-6+\sqrt{31}i+12$$

Add 1\left(-6+i\sqrt{31}\right) to 12.

$$x=6+\sqrt{31}i$$

Now substitute -6-i\sqrt{31} for y in the equation x=y+12 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\sqrt{31}i-6+12$$

Add 1\left(-6-i\sqrt{31}\right) to 12.

$$x=-\sqrt{31}i+6$$

The system is now solved.

$$x=6+\sqrt{31}i,y=-6+\sqrt{31}i\text{ or }x=-\sqrt{31}i+6,y=-\sqrt{31}i-6$$