Solve {l}{sqrt{3}x+y=2}{x^2+y^2=1}

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

Solve for x, y

$$x=\frac{\sqrt{3}}{2}\approx 0.866025404$$

$$y=\frac{1}{2}=0.5$$

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.

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

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

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

Subtract y from both sides of the equation.

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

Divide both sides by \sqrt{3}.

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

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

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

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

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

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

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

Subtract 1 from both sides of the equation.

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

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

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

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

$$y=\frac{-2\left(-\frac{\sqrt{3}}{3}\right)\times \frac{2\sqrt{3}}{3}±\sqrt{\frac{16}{9}-4\left(\left(-\frac{\sqrt{3}}{3}\right)^{2}+1\right)\times \frac{1}{3}}}{2\left(\left(-\frac{\sqrt{3}}{3}\right)^{2}+1\right)}$$

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

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

Multiply -\frac{16}{3} times \frac{1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

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

Add \frac{16}{9} to -\frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Take the square root of 0.

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

The opposite of 1\times 2\left(-\frac{\sqrt{3}}{3}\right)\times \frac{2\sqrt{3}}{3} is \frac{4}{3}.

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

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

$$y=\frac{\frac{4}{3}}{\frac{8}{3}}$$

Divide \frac{4}{3} by \frac{8}{3} by multiplying \frac{4}{3} by the reciprocal of \frac{8}{3}.

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

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

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

Multiply -\frac{\sqrt{3}}{3} times \frac{1}{2}.

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

The system is now solved.

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