Solve {c}{x^2+y^2=2}{2x+y=1}

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

Solve for x, y

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

$$x=-\frac{1}{5}=-0.2\text{, }y=\frac{7}{5}=1.4$$

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.

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

Solve 2x+y=1 for x by isolating x on the left hand side of the equal sign.

$$2x+y=1$$

Subtract y from both sides of the equation.

$$2x=-y+1$$

Divide both sides by 2.

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

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

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

Square -\frac{1}{2}y+\frac{1}{2}.

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

Add y^{2} to \frac{1}{4}y^{2}.

$$\frac{5}{4}y^{2}-\frac{1}{2}y+\frac{1}{4}=2$$

Subtract 2 from both sides of the equation.

$$\frac{5}{4}y^{2}-\frac{1}{2}y-\frac{7}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-\frac{1}{2}\right)^{2} for a, 1\times \frac{1}{2}\left(-\frac{1}{2}\right)\times 2 for b, and -\frac{7}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\times \frac{5}{4}\left(-\frac{7}{4}\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-4\times \frac{5}{4}\left(-\frac{7}{4}\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-5\left(-\frac{7}{4}\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -\frac{7}{4}.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1+35}{4}}}{2\times \frac{5}{4}}$$

Add \frac{1}{4} to \frac{35}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{9}}{2\times \frac{5}{4}}$$

Take the square root of 9.

$$y=\frac{-\left(-\frac{1}{2}\right)±3}{2\times \frac{5}{4}}$$

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

$$y=\frac{\frac{1}{2}±3}{2\times \frac{5}{4}}$$

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

$$y=\frac{\frac{1}{2}±3}{\frac{5}{2}}$$

Now solve the equation y=\frac{\frac{1}{2}±3}{\frac{5}{2}} when ± is plus. Add \frac{1}{2} to 3.

$$y=\frac{\frac{7}{2}}{\frac{5}{2}}$$

Divide \frac{7}{2} by \frac{5}{2} by multiplying \frac{7}{2} by the reciprocal of \frac{5}{2}.

$$y=\frac{7}{5}$$

Now solve the equation y=\frac{\frac{1}{2}±3}{\frac{5}{2}} when ± is minus. Subtract 3 from \frac{1}{2}.

$$y=-\frac{\frac{5}{2}}{\frac{5}{2}}$$

Divide -\frac{5}{2} by \frac{5}{2} by multiplying -\frac{5}{2} by the reciprocal of \frac{5}{2}.

$$y=-1$$

There are two solutions for y: \frac{7}{5} and -1. Substitute \frac{7}{5} for y in the equation x=-\frac{1}{2}y+\frac{1}{2} to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\times \frac{7}{5}+\frac{1}{2}$$

Multiply -\frac{1}{2} times \frac{7}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{7}{10}+\frac{1}{2}$$

Add -\frac{1}{2}\times \frac{7}{5} to \frac{1}{2}.

$$x=-\frac{1}{5}$$

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

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

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

$$x=\frac{1+1}{2}$$

Add -\left(-\frac{1}{2}\right) to \frac{1}{2}.

$$x=1$$

The system is now solved.

$$x=-\frac{1}{5},y=\frac{7}{5}\text{ or }x=1,y=-1$$