Solve {l}{x-y=20}{y^2+x^2=1}

$$\left. \begin{array} { l } { x - y = 20 } \\ { y ^ { 2 } + x ^ { 2 } = 1 } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=-\frac{\sqrt{398}i}{2}+10\approx 10-9.974968672i\text{, }y=-\frac{\sqrt{398}i}{2}-10\approx -10-9.974968672i$$

$$x=\frac{\sqrt{398}i}{2}+10\approx 10+9.974968672i\text{, }y=\frac{\sqrt{398}i}{2}-10\approx -10+9.974968672i$$

Steps Using Substitution and Quadratic Formula

Solve x-y=20 for x by isolating x on the left hand side of the equal sign.

$$x-y=20$$

Subtract -y from both sides of the equation.

$$x=y+20$$

Substitute y+20 for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(y+20\right)^{2}=1$$

Square y+20.

$$y^{2}+y^{2}+40y+400=1$$

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

$$2y^{2}+40y+400=1$$

Subtract 1 from both sides of the equation.

$$2y^{2}+40y+399=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 20\times 1\times 2 for b, and 399 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-40±\sqrt{40^{2}-4\times 2\times 399}}{2\times 2}$$

Square 1\times 20\times 1\times 2.

$$y=\frac{-40±\sqrt{1600-4\times 2\times 399}}{2\times 2}$$

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

$$y=\frac{-40±\sqrt{1600-8\times 399}}{2\times 2}$$

Multiply -8 times 399.

$$y=\frac{-40±\sqrt{1600-3192}}{2\times 2}$$

Add 1600 to -3192.

$$y=\frac{-40±\sqrt{-1592}}{2\times 2}$$

Take the square root of -1592.

$$y=\frac{-40±2\sqrt{398}i}{2\times 2}$$

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

$$y=\frac{-40±2\sqrt{398}i}{4}$$

Now solve the equation y=\frac{-40±2\sqrt{398}i}{4} when ± is plus. Add -40 to 2i\sqrt{398}.

$$y=\frac{-40+2\sqrt{398}i}{4}$$

Divide -40+2i\sqrt{398} by 4.

$$y=\frac{\sqrt{398}i}{2}-10$$

Now solve the equation y=\frac{-40±2\sqrt{398}i}{4} when ± is minus. Subtract 2i\sqrt{398} from -40.

$$y=\frac{-2\sqrt{398}i-40}{4}$$

Divide -40-2i\sqrt{398} by 4.

$$y=-\frac{\sqrt{398}i}{2}-10$$

There are two solutions for y: -10+\frac{i\sqrt{398}}{2} and -10-\frac{i\sqrt{398}}{2}. Substitute -10+\frac{i\sqrt{398}}{2} for y in the equation x=y+20 to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{398}i}{2}-10+20$$

Add 1\left(-10+\frac{i\sqrt{398}}{2}\right) to 20.

$$x=\frac{\sqrt{398}i}{2}+10$$

Now substitute -10-\frac{i\sqrt{398}}{2} for y in the equation x=y+20 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{398}i}{2}-10+20$$

Add 1\left(-10-\frac{i\sqrt{398}}{2}\right) to 20.

$$x=-\frac{\sqrt{398}i}{2}+10$$

The system is now solved.

$$x=\frac{\sqrt{398}i}{2}+10,y=\frac{\sqrt{398}i}{2}-10\text{ or }x=-\frac{\sqrt{398}i}{2}+10,y=-\frac{\sqrt{398}i}{2}-10$$