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

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

Solve for x, y

$$x=\frac{1-\sqrt{3}}{2}\approx -0.366025404\text{, }y=\frac{-\sqrt{3}-1}{2}\approx -1.366025404$$

$$x=\frac{\sqrt{3}+1}{2}\approx 1.366025404\text{, }y=\frac{\sqrt{3}-1}{2}\approx 0.366025404$$

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.

$$x-y=1,y^{2}+x^{2}=2$$

Solve x-y=1 for x by isolating x on the left hand side of the equal sign.

$$x-y=1$$

Subtract -y from both sides of the equation.

$$x=y+1$$

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

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

Square y+1.

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

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

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

Subtract 2 from both sides of the equation.

$$2y^{2}+2y-1=0$$

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

$$y=\frac{-2±\sqrt{2^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

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

$$y=\frac{-2±\sqrt{4-4\times 2\left(-1\right)}}{2\times 2}$$

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

$$y=\frac{-2±\sqrt{4-8\left(-1\right)}}{2\times 2}$$

Multiply -8 times -1.

$$y=\frac{-2±\sqrt{4+8}}{2\times 2}$$

Add 4 to 8.

$$y=\frac{-2±\sqrt{12}}{2\times 2}$$

Take the square root of 12.

$$y=\frac{-2±2\sqrt{3}}{2\times 2}$$

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

$$y=\frac{-2±2\sqrt{3}}{4}$$

Now solve the equation y=\frac{-2±2\sqrt{3}}{4} when ± is plus. Add -2 to 2\sqrt{3}.

$$y=\frac{2\sqrt{3}-2}{4}$$

Divide -2+2\sqrt{3} by 4.

$$y=\frac{\sqrt{3}-1}{2}$$

Now solve the equation y=\frac{-2±2\sqrt{3}}{4} when ± is minus. Subtract 2\sqrt{3} from -2.

$$y=\frac{-2\sqrt{3}-2}{4}$$

Divide -2-2\sqrt{3} by 4.

$$y=\frac{-\sqrt{3}-1}{2}$$

There are two solutions for y: \frac{-1+\sqrt{3}}{2} and \frac{-1-\sqrt{3}}{2}. Substitute \frac{-1+\sqrt{3}}{2} for y in the equation x=y+1 to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{3}-1}{2}+1$$

Now substitute \frac{-1-\sqrt{3}}{2} for y in the equation x=y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{-\sqrt{3}-1}{2}+1$$

The system is now solved.

$$x=\frac{\sqrt{3}-1}{2}+1,y=\frac{\sqrt{3}-1}{2}\text{ or }x=\frac{-\sqrt{3}-1}{2}+1,y=\frac{-\sqrt{3}-1}{2}$$