Solve {c}{1/2x+frac{sqrt{3}}{2}y=-frac{sqrt{3}}{3}}{x^2+y^2=1}

$$\left. \begin{array} { c } { \frac { 1 } { 2 } x + \frac { \sqrt { 3 } } { 2 } y = - \frac { \sqrt { 3 } } { 3 } } \\ { x ^ { 2 } + y ^ { 2 } = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{3\sqrt{2}-\sqrt{3}}{6}\approx 0.418431647\text{, }y=-\frac{\sqrt{6}}{6}-\frac{1}{2}\approx -0.90824829$$

$$x=-\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{6}\approx -0.995781916\text{, }y=\frac{\sqrt{6}}{6}-\frac{1}{2}\approx -0.09175171$$

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.

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

Solve \frac{1}{2}x+\frac{\sqrt{3}}{2}y=-\frac{\sqrt{3}}{3} for x by isolating x on the left hand side of the equal sign.

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

Subtract \frac{\sqrt{3}}{2}y from both sides of the equation.

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

Multiply both sides by 2.

$$x=\left(-\sqrt{3}\right)y-\frac{2\sqrt{3}}{3}$$

Substitute \left(-\sqrt{3}\right)y-\frac{2\sqrt{3}}{3} for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(\left(-\sqrt{3}\right)y-\frac{2\sqrt{3}}{3}\right)^{2}=1$$

Square \left(-\sqrt{3}\right)y-\frac{2\sqrt{3}}{3}.

$$y^{2}+\left(-\sqrt{3}\right)^{2}y^{2}+2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)y+\left(-\frac{2\sqrt{3}}{3}\right)^{2}=1$$

Add y^{2} to \left(-\sqrt{3}\right)^{2}y^{2}.

$$\left(\left(-\sqrt{3}\right)^{2}+1\right)y^{2}+2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)y+\left(-\frac{2\sqrt{3}}{3}\right)^{2}=1$$

Subtract 1 from both sides of the equation.

$$\left(\left(-\sqrt{3}\right)^{2}+1\right)y^{2}+2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)y+\left(-\frac{2\sqrt{3}}{3}\right)^{2}-1=0$$

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

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

Square 1\times 2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right).

$$y=\frac{-2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)±\sqrt{16-4\left(\left(-\sqrt{3}\right)^{2}+1\right)\times \frac{1}{3}}}{2\left(\left(-\sqrt{3}\right)^{2}+1\right)}$$

Multiply -4 times 1+1\left(-\sqrt{3}\right)^{2}.

$$y=\frac{-2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)±\sqrt{16-16\times \frac{1}{3}}}{2\left(\left(-\sqrt{3}\right)^{2}+1\right)}$$

Multiply -16 times \frac{1}{3}.

$$y=\frac{-2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)±\sqrt{16-\frac{16}{3}}}{2\left(\left(-\sqrt{3}\right)^{2}+1\right)}$$

Add 16 to -\frac{16}{3}.

$$y=\frac{-2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)±\sqrt{\frac{32}{3}}}{2\left(\left(-\sqrt{3}\right)^{2}+1\right)}$$

Take the square root of \frac{32}{3}.

$$y=\frac{-2\left(-\sqrt{3}\right)\left(-\frac{2\sqrt{3}}{3}\right)±\frac{4\sqrt{6}}{3}}{2\left(\left(-\sqrt{3}\right)^{2}+1\right)}$$

Multiply 2 times 1+1\left(-\sqrt{3}\right)^{2}.

$$y=\frac{-4±\frac{4\sqrt{6}}{3}}{8}$$

Now solve the equation y=\frac{-4±\frac{4\sqrt{6}}{3}}{8} when ± is plus. Add -4 to \frac{4\sqrt{6}}{3}.

$$y=\frac{\frac{4\sqrt{6}}{3}-4}{8}$$

Divide -4+\frac{4\sqrt{6}}{3} by 8.

$$y=\frac{\sqrt{6}}{6}-\frac{1}{2}$$

Now solve the equation y=\frac{-4±\frac{4\sqrt{6}}{3}}{8} when ± is minus. Subtract \frac{4\sqrt{6}}{3} from -4.

$$y=\frac{-\frac{4\sqrt{6}}{3}-4}{8}$$

Divide -4-\frac{4\sqrt{6}}{3} by 8.

$$y=-\frac{\sqrt{6}}{6}-\frac{1}{2}$$

There are two solutions for y: -\frac{1}{2}+\frac{\sqrt{6}}{6} and -\frac{1}{2}-\frac{\sqrt{6}}{6}. Substitute -\frac{1}{2}+\frac{\sqrt{6}}{6} for y in the equation x=\left(-\sqrt{3}\right)y-\frac{2\sqrt{3}}{3} to find the corresponding solution for x that satisfies both equations.

$$x=\left(-\sqrt{3}\right)\left(\frac{\sqrt{6}}{6}-\frac{1}{2}\right)-\frac{2\sqrt{3}}{3}$$

Now substitute -\frac{1}{2}-\frac{\sqrt{6}}{6} for y in the equation x=\left(-\sqrt{3}\right)y-\frac{2\sqrt{3}}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\left(-\sqrt{3}\right)\left(-\frac{\sqrt{6}}{6}-\frac{1}{2}\right)-\frac{2\sqrt{3}}{3}$$

The system is now solved.

$$x=\left(-\sqrt{3}\right)\left(\frac{\sqrt{6}}{6}-\frac{1}{2}\right)-\frac{2\sqrt{3}}{3},y=\frac{\sqrt{6}}{6}-\frac{1}{2}\text{ or }x=\left(-\sqrt{3}\right)\left(-\frac{\sqrt{6}}{6}-\frac{1}{2}\right)-\frac{2\sqrt{3}}{3},y=-\frac{\sqrt{6}}{6}-\frac{1}{2}$$