Solve {l}{2x+y=5}{x^2+y^2=25}

$$\left. \begin{array} { l } { 2 x + y = 5 } \\ { x ^ { 2 } + y ^ { 2 } = 25 } \end{array} \right.$$

Solve for x, y

$$x=0\text{, }y=5$$

$$x=4\text{, }y=-3$$

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.

$$2x+y=5,y^{2}+x^{2}=25$$

Solve 2x+y=5 for x by isolating x on the left hand side of the equal sign.

$$2x+y=5$$

Subtract y from both sides of the equation.

$$2x=-y+5$$

Divide both sides by 2.

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

Substitute -\frac{1}{2}y+\frac{5}{2} for x in the other equation, y^{2}+x^{2}=25.

$$y^{2}+\left(-\frac{1}{2}y+\frac{5}{2}\right)^{2}=25$$

Square -\frac{1}{2}y+\frac{5}{2}.

$$y^{2}+\frac{1}{4}y^{2}-\frac{5}{2}y+\frac{25}{4}=25$$

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

$$\frac{5}{4}y^{2}-\frac{5}{2}y+\frac{25}{4}=25$$

Subtract 25 from both sides of the equation.

$$\frac{5}{4}y^{2}-\frac{5}{2}y-\frac{75}{4}=0$$

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{5}{2}\left(-\frac{1}{2}\right)\times 2 for b, and -\frac{75}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\left(-\frac{5}{2}\right)^{2}-4\times \frac{5}{4}\left(-\frac{75}{4}\right)}}{2\times \frac{5}{4}}$$

Square 1\times \frac{5}{2}\left(-\frac{1}{2}\right)\times 2.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-4\times \frac{5}{4}\left(-\frac{75}{4}\right)}}{2\times \frac{5}{4}}$$

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

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25}{4}-5\left(-\frac{75}{4}\right)}}{2\times \frac{5}{4}}$$

Multiply -5 times -\frac{75}{4}.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{\frac{25+375}{4}}}{2\times \frac{5}{4}}$$

Add \frac{25}{4} to \frac{375}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{5}{2}\right)±\sqrt{100}}{2\times \frac{5}{4}}$$

Take the square root of 100.

$$y=\frac{-\left(-\frac{5}{2}\right)±10}{2\times \frac{5}{4}}$$

The opposite of 1\times \frac{5}{2}\left(-\frac{1}{2}\right)\times 2 is \frac{5}{2}.

$$y=\frac{\frac{5}{2}±10}{2\times \frac{5}{4}}$$

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

$$y=\frac{\frac{5}{2}±10}{\frac{5}{2}}$$

Now solve the equation y=\frac{\frac{5}{2}±10}{\frac{5}{2}} when ± is plus. Add \frac{5}{2} to 10.

$$y=\frac{\frac{25}{2}}{\frac{5}{2}}$$

Divide \frac{25}{2} by \frac{5}{2} by multiplying \frac{25}{2} by the reciprocal of \frac{5}{2}.

$$y=5$$

Now solve the equation y=\frac{\frac{5}{2}±10}{\frac{5}{2}} when ± is minus. Subtract 10 from \frac{5}{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: 5 and -3. Substitute 5 for y in the equation x=-\frac{1}{2}y+\frac{5}{2} to find the corresponding solution for x that satisfies both equations.

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

Multiply -\frac{1}{2} times 5.

$$x=\frac{-5+5}{2}$$

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

$$x=0$$

Now substitute -3 for y in the equation x=-\frac{1}{2}y+\frac{5}{2} and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{1}{2}\left(-3\right)+\frac{5}{2}$$

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

$$x=\frac{3+5}{2}$$

Add -3\left(-\frac{1}{2}\right) to \frac{5}{2}.

$$x=4$$

The system is now solved.

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