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

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

Solve for x, y

$$x=\frac{1-\sqrt{15}}{4}\approx -0.718245837\text{, }y=\frac{-\sqrt{15}-1}{4}\approx -1.218245837$$

$$x=\frac{\sqrt{15}+1}{4}\approx 1.218245837\text{, }y=\frac{\sqrt{15}-1}{4}\approx 0.718245837$$

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.

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

Solve x-y=\frac{1}{2} for x by isolating x on the left hand side of the equal sign.

$$x-y=\frac{1}{2}$$

Subtract -y from both sides of the equation.

$$x=y+\frac{1}{2}$$

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

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

Square y+\frac{1}{2}.

$$y^{2}+y^{2}+y+\frac{1}{4}=2$$

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

$$2y^{2}+y+\frac{1}{4}=2$$

Subtract 2 from both sides of the equation.

$$2y^{2}+y-\frac{7}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times \frac{1}{2}\times 1\times 2 for b, and -\frac{7}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-1±\sqrt{1^{2}-4\times 2\left(-\frac{7}{4}\right)}}{2\times 2}$$

Square 1\times \frac{1}{2}\times 1\times 2.

$$y=\frac{-1±\sqrt{1-4\times 2\left(-\frac{7}{4}\right)}}{2\times 2}$$

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

$$y=\frac{-1±\sqrt{1-8\left(-\frac{7}{4}\right)}}{2\times 2}$$

Multiply -8 times -\frac{7}{4}.

$$y=\frac{-1±\sqrt{1+14}}{2\times 2}$$

Add 1 to 14.

$$y=\frac{-1±\sqrt{15}}{2\times 2}$$

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

$$y=\frac{-1±\sqrt{15}}{4}$$

Now solve the equation y=\frac{-1±\sqrt{15}}{4} when ± is plus. Add -1 to \sqrt{15}.

$$y=\frac{\sqrt{15}-1}{4}$$

Now solve the equation y=\frac{-1±\sqrt{15}}{4} when ± is minus. Subtract \sqrt{15} from -1.

$$y=\frac{-\sqrt{15}-1}{4}$$

Both solutions for y are the same: \frac{-1+\sqrt{15}}{4}. Substitute \frac{-1+\sqrt{15}}{4} for y in the equation x=y+\frac{1}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{15}-1}{4}+\frac{1}{2}$$

Now substitute \frac{-1-\sqrt{15}}{4} for y in the equation x=y+\frac{1}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{-\sqrt{15}-1}{4}+\frac{1}{2}$$

The system is now solved.

$$x=\frac{\sqrt{15}-1}{4}+\frac{1}{2},y=\frac{\sqrt{15}-1}{4}\text{ or }x=\frac{-\sqrt{15}-1}{4}+\frac{1}{2},y=\frac{-\sqrt{15}-1}{4}$$