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

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

Solve for x, y

$$x=\frac{\sqrt{14}}{6}+\frac{1}{3}\approx 0.956942898\text{, }y=-\frac{\sqrt{14}}{6}+\frac{1}{3}\approx -0.290276231$$

$$x=-\frac{\sqrt{14}}{6}+\frac{1}{3}\approx -0.290276231\text{, }y=\frac{\sqrt{14}}{6}+\frac{1}{3}\approx 0.956942898$$

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.

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

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

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

Subtract y from both sides of the equation.

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

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

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

Square -y+\frac{2}{3}.

$$y^{2}+y^{2}-\frac{4}{3}y+\frac{4}{9}=1$$

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

$$2y^{2}-\frac{4}{3}y+\frac{4}{9}=1$$

Subtract 1 from both sides of the equation.

$$2y^{2}-\frac{4}{3}y-\frac{5}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times \frac{2}{3}\left(-1\right)\times 2 for b, and -\frac{5}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times 2\left(-\frac{5}{9}\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16}{9}-4\times 2\left(-\frac{5}{9}\right)}}{2\times 2}$$

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

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16}{9}-8\left(-\frac{5}{9}\right)}}{2\times 2}$$

Multiply -8 times -\frac{5}{9}.

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{16+40}{9}}}{2\times 2}$$

Add \frac{16}{9} to \frac{40}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\frac{56}{9}}}{2\times 2}$$

Take the square root of \frac{56}{9}.

$$y=\frac{-\left(-\frac{4}{3}\right)±\frac{2\sqrt{14}}{3}}{2\times 2}$$

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

$$y=\frac{\frac{4}{3}±\frac{2\sqrt{14}}{3}}{2\times 2}$$

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

$$y=\frac{\frac{4}{3}±\frac{2\sqrt{14}}{3}}{4}$$

Now solve the equation y=\frac{\frac{4}{3}±\frac{2\sqrt{14}}{3}}{4} when ± is plus. Add \frac{4}{3} to \frac{2\sqrt{14}}{3}.

$$y=\frac{2\sqrt{14}+4}{3\times 4}$$

Divide \frac{4+2\sqrt{14}}{3} by 4.

$$y=\frac{\sqrt{14}}{6}+\frac{1}{3}$$

Now solve the equation y=\frac{\frac{4}{3}±\frac{2\sqrt{14}}{3}}{4} when ± is minus. Subtract \frac{2\sqrt{14}}{3} from \frac{4}{3}.

$$y=\frac{4-2\sqrt{14}}{3\times 4}$$

Divide \frac{4-2\sqrt{14}}{3} by 4.

$$y=-\frac{\sqrt{14}}{6}+\frac{1}{3}$$

There are two solutions for y: \frac{1}{3}+\frac{\sqrt{14}}{6} and \frac{1}{3}-\frac{\sqrt{14}}{6}. Substitute \frac{1}{3}+\frac{\sqrt{14}}{6} for y in the equation x=-y+\frac{2}{3} to find the corresponding solution for x that satisfies both equations.

$$x=-\left(\frac{\sqrt{14}}{6}+\frac{1}{3}\right)+\frac{2}{3}$$

Now substitute \frac{1}{3}-\frac{\sqrt{14}}{6} for y in the equation x=-y+\frac{2}{3} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\left(-\frac{\sqrt{14}}{6}+\frac{1}{3}\right)+\frac{2}{3}$$

The system is now solved.

$$x=-\left(\frac{\sqrt{14}}{6}+\frac{1}{3}\right)+\frac{2}{3},y=\frac{\sqrt{14}}{6}+\frac{1}{3}\text{ or }x=-\left(-\frac{\sqrt{14}}{6}+\frac{1}{3}\right)+\frac{2}{3},y=-\frac{\sqrt{14}}{6}+\frac{1}{3}$$