Solve {l}{x^2+y^2=50}{2x-y=13}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 50 } \\ { 2 x - y = 13 } \end{array} \right.$$

Solve for x, y

$$x=\frac{17}{5}=3.4\text{, }y=-\frac{31}{5}=-6.2$$

$$x=7\text{, }y=1$$

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.

$$2x-y=13,y^{2}+x^{2}=50$$

Solve 2x-y=13 for x by isolating x on the left hand side of the equal sign.

$$2x-y=13$$

Subtract -y from both sides of the equation.

$$2x=y+13$$

Divide both sides by 2.

$$x=\frac{1}{2}y+\frac{13}{2}$$

Substitute \frac{1}{2}y+\frac{13}{2} for x in the other equation, y^{2}+x^{2}=50.

$$y^{2}+\left(\frac{1}{2}y+\frac{13}{2}\right)^{2}=50$$

Square \frac{1}{2}y+\frac{13}{2}.

$$y^{2}+\frac{1}{4}y^{2}+\frac{13}{2}y+\frac{169}{4}=50$$

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

$$\frac{5}{4}y^{2}+\frac{13}{2}y+\frac{169}{4}=50$$

Subtract 50 from both sides of the equation.

$$\frac{5}{4}y^{2}+\frac{13}{2}y-\frac{31}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{1}{2}\right)^{2} for a, 1\times \frac{13}{2}\times \frac{1}{2}\times 2 for b, and -\frac{31}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\frac{13}{2}±\sqrt{\left(\frac{13}{2}\right)^{2}-4\times \frac{5}{4}\left(-\frac{31}{4}\right)}}{2\times \frac{5}{4}}$$

Square 1\times \frac{13}{2}\times \frac{1}{2}\times 2.

$$y=\frac{-\frac{13}{2}±\sqrt{\frac{169}{4}-4\times \frac{5}{4}\left(-\frac{31}{4}\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\frac{13}{2}±\sqrt{\frac{169}{4}-5\left(-\frac{31}{4}\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -\frac{31}{4}.

$$y=\frac{-\frac{13}{2}±\sqrt{\frac{169+155}{4}}}{2\times \frac{5}{4}}$$

Add \frac{169}{4} to \frac{155}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\frac{13}{2}±\sqrt{81}}{2\times \frac{5}{4}}$$

Take the square root of 81.

$$y=\frac{-\frac{13}{2}±9}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\frac{13}{2}±9}{\frac{5}{2}}$$

Now solve the equation y=\frac{-\frac{13}{2}±9}{\frac{5}{2}} when ± is plus. Add -\frac{13}{2} to 9.

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

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

$$y=1$$

Now solve the equation y=\frac{-\frac{13}{2}±9}{\frac{5}{2}} when ± is minus. Subtract 9 from -\frac{13}{2}.

$$y=-\frac{\frac{31}{2}}{\frac{5}{2}}$$

Divide -\frac{31}{2} by \frac{5}{2} by multiplying -\frac{31}{2} by the reciprocal of \frac{5}{2}.

$$y=-\frac{31}{5}$$

There are two solutions for y: 1 and -\frac{31}{5}. Substitute 1 for y in the equation x=\frac{1}{2}y+\frac{13}{2} to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1+13}{2}$$

Add \frac{1}{2}\times 1 to \frac{13}{2}.

$$x=7$$

Now substitute -\frac{31}{5} for y in the equation x=\frac{1}{2}y+\frac{13}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\left(-\frac{31}{5}\right)+\frac{13}{2}$$

Multiply \frac{1}{2} times -\frac{31}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{31}{10}+\frac{13}{2}$$

Add -\frac{31}{5}\times \frac{1}{2} to \frac{13}{2}.

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

The system is now solved.

$$x=7,y=1\text{ or }x=\frac{17}{5},y=-\frac{31}{5}$$