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

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

Solve for x, y

$$x=0\text{, }y=-1$$

$$x=\frac{3}{5}=0.6\text{, }y=\frac{4}{5}=0.8$$

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.

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

Solve 3x-y=1 for x by isolating x on the left hand side of the equal sign.

$$3x-y=1$$

Subtract -y from both sides of the equation.

$$3x=y+1$$

Divide both sides by 3.

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

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

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

Square \frac{1}{3}y+\frac{1}{3}.

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

Add y^{2} to \frac{1}{9}y^{2}.

$$\frac{10}{9}y^{2}+\frac{2}{9}y+\frac{1}{9}=1$$

Subtract 1 from both sides of the equation.

$$\frac{10}{9}y^{2}+\frac{2}{9}y-\frac{8}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{1}{3}\right)^{2} for a, 1\times \frac{1}{3}\times \frac{1}{3}\times 2 for b, and -\frac{8}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{2}{9}±\sqrt{\left(\frac{2}{9}\right)^{2}-4\times \frac{10}{9}\left(-\frac{8}{9}\right)}}{2\times \frac{10}{9}}$$

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

$$y=\frac{-\frac{2}{9}±\sqrt{\frac{4}{81}-4\times \frac{10}{9}\left(-\frac{8}{9}\right)}}{2\times \frac{10}{9}}$$

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

$$y=\frac{-\frac{2}{9}±\sqrt{\frac{4}{81}-\frac{40}{9}\left(-\frac{8}{9}\right)}}{2\times \frac{10}{9}}$$

Multiply -\frac{40}{9} times -\frac{8}{9} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{2}{9}±\sqrt{\frac{4+320}{81}}}{2\times \frac{10}{9}}$$

Add \frac{4}{81} to \frac{320}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{2}{9}±\sqrt{4}}{2\times \frac{10}{9}}$$

Take the square root of 4.

$$y=\frac{-\frac{2}{9}±2}{2\times \frac{10}{9}}$$

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

$$y=\frac{-\frac{2}{9}±2}{\frac{20}{9}}$$

Now solve the equation y=\frac{-\frac{2}{9}±2}{\frac{20}{9}} when ± is plus. Add -\frac{2}{9} to 2.

$$y=\frac{\frac{16}{9}}{\frac{20}{9}}$$

Divide \frac{16}{9} by \frac{20}{9} by multiplying \frac{16}{9} by the reciprocal of \frac{20}{9}.

$$y=\frac{4}{5}$$

Now solve the equation y=\frac{-\frac{2}{9}±2}{\frac{20}{9}} when ± is minus. Subtract 2 from -\frac{2}{9}.

$$y=-\frac{\frac{20}{9}}{\frac{20}{9}}$$

Divide -\frac{20}{9} by \frac{20}{9} by multiplying -\frac{20}{9} by the reciprocal of \frac{20}{9}.

$$y=-1$$

There are two solutions for y: \frac{4}{5} and -1. Substitute \frac{4}{5} for y in the equation x=\frac{1}{3}y+\frac{1}{3} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{3}\times \frac{4}{5}+\frac{1}{3}$$

Multiply \frac{1}{3} times \frac{4}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{4}{15}+\frac{1}{3}$$

Add \frac{1}{3}\times \frac{4}{5} to \frac{1}{3}.

$$x=\frac{3}{5}$$

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

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

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

$$x=\frac{-1+1}{3}$$

Add -\frac{1}{3} to \frac{1}{3}.

$$x=0$$

The system is now solved.

$$x=\frac{3}{5},y=\frac{4}{5}\text{ or }x=0,y=-1$$