Solve {l}{x^2+y^2=1}{x+y=0.5}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { x + y = 0.5 } \end{array} \right.$$

Solve for x, y

$$x=\frac{\sqrt{7}+1}{4}\approx 0.911437828\text{, }y=\frac{1-\sqrt{7}}{4}\approx -0.411437828$$

$$x=\frac{1-\sqrt{7}}{4}\approx -0.411437828\text{, }y=\frac{\sqrt{7}+1}{4}\approx 0.911437828$$

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.

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

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}=1.

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

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

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

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

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

Subtract 1 from both sides of the equation.

$$2y^{2}-y-\frac{3}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times \frac{1}{2}\left(-1\right)\times 2 for b, and -\frac{3}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-\frac{3}{4}\right)}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$y=\frac{-\left(-1\right)±\sqrt{1-8\left(-\frac{3}{4}\right)}}{2\times 2}$$

Multiply -8 times -\frac{3}{4}.

$$y=\frac{-\left(-1\right)±\sqrt{1+6}}{2\times 2}$$

Add 1 to 6.

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

The opposite of 1\times \frac{1}{2}\left(-1\right)\times 2 is 1.

$$y=\frac{1±\sqrt{7}}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$y=\frac{1±\sqrt{7}}{4}$$

Now solve the equation y=\frac{1±\sqrt{7}}{4} when ± is plus. Add 1 to \sqrt{7}.

$$y=\frac{\sqrt{7}+1}{4}$$

Now solve the equation y=\frac{1±\sqrt{7}}{4} when ± is minus. Subtract \sqrt{7} from 1.

$$y=\frac{1-\sqrt{7}}{4}$$

There are two solutions for y: \frac{1+\sqrt{7}}{4} and \frac{1-\sqrt{7}}{4}. Substitute \frac{1+\sqrt{7}}{4} for y in the equation x=-y+\frac{1}{2} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{\sqrt{7}+1}{4}+\frac{1}{2}$$

Now substitute \frac{1-\sqrt{7}}{4} for y in the equation x=-y+\frac{1}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1-\sqrt{7}}{4}+\frac{1}{2}$$

The system is now solved.

$$x=-\frac{\sqrt{7}+1}{4}+\frac{1}{2},y=\frac{\sqrt{7}+1}{4}\text{ or }x=-\frac{1-\sqrt{7}}{4}+\frac{1}{2},y=\frac{1-\sqrt{7}}{4}$$