Solve {l}{2x+y=1}{x^2+y^2=13}

$$\left\{ \begin{array} { l } { 2 x + y = 1 } \\ { x ^ { 2 } + y ^ { 2 } = 13 } \end{array} \right.$$

Solve for x, y

$$x=2\text{, }y=-3$$

$$x=-\frac{6}{5}=-1.2\text{, }y=\frac{17}{5}=3.4$$

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.

$$2x+y=1,y^{2}+x^{2}=13$$

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}=13.

$$y^{2}+\left(-\frac{1}{2}y+\frac{1}{2}\right)^{2}=13$$

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

$$y^{2}+\frac{1}{4}y^{2}-\frac{1}{2}y+\frac{1}{4}=13$$

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

$$\frac{5}{4}y^{2}-\frac{1}{2}y+\frac{1}{4}=13$$

Subtract 13 from both sides of the equation.

$$\frac{5}{4}y^{2}-\frac{1}{2}y-\frac{51}{4}=0$$

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{51}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\times \frac{5}{4}\left(-\frac{51}{4}\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-4\times \frac{5}{4}\left(-\frac{51}{4}\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1}{4}-5\left(-\frac{51}{4}\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -\frac{51}{4}.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\frac{1+255}{4}}}{2\times \frac{5}{4}}$$

Add \frac{1}{4} to \frac{255}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{1}{2}\right)±\sqrt{64}}{2\times \frac{5}{4}}$$

Take the square root of 64.

$$y=\frac{-\left(-\frac{1}{2}\right)±8}{2\times \frac{5}{4}}$$

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

$$y=\frac{\frac{1}{2}±8}{2\times \frac{5}{4}}$$

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

$$y=\frac{\frac{1}{2}±8}{\frac{5}{2}}$$

Now solve the equation y=\frac{\frac{1}{2}±8}{\frac{5}{2}} when ± is plus. Add \frac{1}{2} to 8.

$$y=\frac{\frac{17}{2}}{\frac{5}{2}}$$

Divide \frac{17}{2} by \frac{5}{2} by multiplying \frac{17}{2} by the reciprocal of \frac{5}{2}.

$$y=\frac{17}{5}$$

Now solve the equation y=\frac{\frac{1}{2}±8}{\frac{5}{2}} when ± is minus. Subtract 8 from \frac{1}{2}.

$$y=-\frac{\frac{15}{2}}{\frac{5}{2}}$$

Divide -\frac{15}{2} by \frac{5}{2} by multiplying -\frac{15}{2} by the reciprocal of \frac{5}{2}.

$$y=-3$$

There are two solutions for y: \frac{17}{5} and -3. Substitute \frac{17}{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.

$$x=-\frac{1}{2}\times \frac{17}{5}+\frac{1}{2}$$

Multiply -\frac{1}{2} times \frac{17}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{17}{10}+\frac{1}{2}$$

Add -\frac{1}{2}\times \frac{17}{5} to \frac{1}{2}.

$$x=-\frac{6}{5}$$

Now substitute -3 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.

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

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

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

Add -3\left(-\frac{1}{2}\right) to \frac{1}{2}.

$$x=2$$

The system is now solved.

$$x=-\frac{6}{5},y=\frac{17}{5}\text{ or }x=2,y=-3$$