Solve {l}{x^2+y^2=6}{x-y=2}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 6 } \\ { x - y = 2 } \end{array} \right.$$

Solve for x, y

$$x=1-\sqrt{2}\approx -0.414213562\text{, }y=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

$$x=\sqrt{2}+1\approx 2.414213562\text{, }y=\sqrt{2}-1\approx 0.414213562$$

Steps Using Substitution and Quadratic Formula

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$x-y=2,y^{2}+x^{2}=6$$

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

$$x-y=2$$

Subtract -y from both sides of the equation.

$$x=y+2$$

Substitute y+2 for x in the other equation, y^{2}+x^{2}=6.

$$y^{2}+\left(y+2\right)^{2}=6$$

Square y+2.

$$y^{2}+y^{2}+4y+4=6$$

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

$$2y^{2}+4y+4=6$$

Subtract 6 from both sides of the equation.

$$2y^{2}+4y-2=0$$

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

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

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

$$y=\frac{-4±\sqrt{16-4\times 2\left(-2\right)}}{2\times 2}$$

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

$$y=\frac{-4±\sqrt{16-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$y=\frac{-4±\sqrt{16+16}}{2\times 2}$$

Add 16 to 16.

$$y=\frac{-4±\sqrt{32}}{2\times 2}$$

Take the square root of 32.

$$y=\frac{-4±4\sqrt{2}}{2\times 2}$$

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

$$y=\frac{-4±4\sqrt{2}}{4}$$

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

$$y=\frac{4\sqrt{2}-4}{4}$$

Divide -4+4\sqrt{2} by 4.

$$y=\sqrt{2}-1$$

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

$$y=\frac{-4\sqrt{2}-4}{4}$$

Divide -4-4\sqrt{2} by 4.

$$y=-\sqrt{2}-1$$

There are two solutions for y: -1+\sqrt{2} and -1-\sqrt{2}. Substitute -1+\sqrt{2} for y in the equation x=y+2 to find the corresponding solution for x that satisfies both equations.

$$x=\sqrt{2}-1+2$$

Add 1\left(-1+\sqrt{2}\right) to 2.

$$x=\sqrt{2}+1$$

Now substitute -1-\sqrt{2} for y in the equation x=y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\sqrt{2}-1+2$$

Add 1\left(-1-\sqrt{2}\right) to 2.

$$x=1-\sqrt{2}$$

The system is now solved.

$$x=\sqrt{2}+1,y=\sqrt{2}-1\text{ or }x=1-\sqrt{2},y=-\sqrt{2}-1$$