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

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

Solve for x, y

$$x=\frac{\sqrt{3}-\sqrt{15}}{8}\approx -0.267616567\text{, }y=\frac{-3\sqrt{5}-1}{8}\approx -0.963525492$$

$$x=\frac{\sqrt{3}+\sqrt{15}}{8}\approx 0.700629269\text{, }y=\frac{3\sqrt{5}-1}{8}\approx 0.713525492$$

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.

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

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

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

Subtract -y from both sides of the equation.

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

Divide both sides by \sqrt{3}.

$$x=\frac{\sqrt{3}}{3}y+\frac{\sqrt{3}}{6}$$

Substitute \frac{\sqrt{3}}{3}y+\frac{\sqrt{3}}{6} for x in the other equation, y^{2}+x^{2}=1.

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

Square \frac{\sqrt{3}}{3}y+\frac{\sqrt{3}}{6}.

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

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

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

Subtract 1 from both sides of the equation.

$$\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)y^{2}+2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}y+\left(\frac{\sqrt{3}}{6}\right)^{2}-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{\sqrt{3}}{3}\right)^{2} for a, 1\times 2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6} for b, and -\frac{11}{12} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}±\sqrt{\left(2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}\right)^{2}-4\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)\left(-\frac{11}{12}\right)}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)}$$

Square 1\times 2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}.

$$y=\frac{-2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}±\sqrt{\frac{1}{9}-4\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)\left(-\frac{11}{12}\right)}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)}$$

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

$$y=\frac{-2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}±\sqrt{\frac{1}{9}-\frac{16}{3}\left(-\frac{11}{12}\right)}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)}$$

Multiply -\frac{16}{3} times -\frac{11}{12} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}±\sqrt{\frac{1+44}{9}}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)}$$

Add \frac{1}{9} to \frac{44}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{6}±\sqrt{5}}{2\left(\left(\frac{\sqrt{3}}{3}\right)^{2}+1\right)}$$

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

$$y=\frac{-\frac{1}{3}±\sqrt{5}}{\frac{8}{3}}$$

Now solve the equation y=\frac{-\frac{1}{3}±\sqrt{5}}{\frac{8}{3}} when ± is plus. Add -\frac{1}{3} to \sqrt{5}.

$$y=\frac{\sqrt{5}-\frac{1}{3}}{\frac{8}{3}}$$

Divide -\frac{1}{3}+\sqrt{5} by \frac{8}{3} by multiplying -\frac{1}{3}+\sqrt{5} by the reciprocal of \frac{8}{3}.

$$y=\frac{3\sqrt{5}-1}{8}$$

Now solve the equation y=\frac{-\frac{1}{3}±\sqrt{5}}{\frac{8}{3}} when ± is minus. Subtract \sqrt{5} from -\frac{1}{3}.

$$y=\frac{-\sqrt{5}-\frac{1}{3}}{\frac{8}{3}}$$

Divide -\frac{1}{3}-\sqrt{5} by \frac{8}{3} by multiplying -\frac{1}{3}-\sqrt{5} by the reciprocal of \frac{8}{3}.

$$y=\frac{-3\sqrt{5}-1}{8}$$

There are two solutions for y: \frac{-1+3\sqrt{5}}{8} and \frac{-1-3\sqrt{5}}{8}. Substitute \frac{-1+3\sqrt{5}}{8} for y in the equation x=\frac{\sqrt{3}}{3}y+\frac{\sqrt{3}}{6} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{3}}{3}\times \frac{3\sqrt{5}-1}{8}+\frac{\sqrt{3}}{6}$$

Now substitute \frac{-1-3\sqrt{5}}{8} for y in the equation x=\frac{\sqrt{3}}{3}y+\frac{\sqrt{3}}{6} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{\sqrt{3}}{3}\times \frac{-3\sqrt{5}-1}{8}+\frac{\sqrt{3}}{6}$$

The system is now solved.

$$x=\frac{\sqrt{3}}{3}\times \frac{3\sqrt{5}-1}{8}+\frac{\sqrt{3}}{6},y=\frac{3\sqrt{5}-1}{8}\text{ or }x=\frac{\sqrt{3}}{3}\times \frac{-3\sqrt{5}-1}{8}+\frac{\sqrt{3}}{6},y=\frac{-3\sqrt{5}-1}{8}$$