Solve {l}{x^2+y^2=13}{2x-y=4}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 13 } \\ { 2 x - y = 4 } \end{array} \right.$$

Solve for x, y

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

$$x=\frac{1}{5}=0.2\text{, }y=-\frac{18}{5}=-3.6$$

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.

$$2x-y=4,y^{2}+x^{2}=13$$

Solve 2x-y=4 for x by isolating x on the left hand side of the equal sign.

$$2x-y=4$$

Subtract -y from both sides of the equation.

$$2x=y+4$$

Divide both sides by 2.

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

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

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

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

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

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

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

Subtract 13 from both sides of the equation.

$$\frac{5}{4}y^{2}+2y-9=0$$

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 2\times \frac{1}{2}\times 2 for b, and -9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

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

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

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

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

Multiply -5 times -9.

$$y=\frac{-2±\sqrt{4+45}}{2\times \frac{5}{4}}$$

Add 4 to 45.

$$y=\frac{-2±\sqrt{49}}{2\times \frac{5}{4}}$$

Take the square root of 49.

$$y=\frac{-2±7}{2\times \frac{5}{4}}$$

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

$$y=\frac{-2±7}{\frac{5}{2}}$$

Now solve the equation y=\frac{-2±7}{\frac{5}{2}} when ± is plus. Add -2 to 7.

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

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

$$y=2$$

Now solve the equation y=\frac{-2±7}{\frac{5}{2}} when ± is minus. Subtract 7 from -2.

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

Divide -9 by \frac{5}{2} by multiplying -9 by the reciprocal of \frac{5}{2}.

$$y=-\frac{18}{5}$$

There are two solutions for y: 2 and -\frac{18}{5}. Substitute 2 for y in the equation x=\frac{1}{2}y+2 to find the corresponding solution for x that satisfies both equations.

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

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

$$x=1+2$$

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

$$x=3$$

Now substitute -\frac{18}{5} for y in the equation x=\frac{1}{2}y+2 and solve to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{2}\left(-\frac{18}{5}\right)+2$$

Multiply \frac{1}{2} times -\frac{18}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{9}{5}+2$$

Add -\frac{18}{5}\times \frac{1}{2} to 2.

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

The system is now solved.

$$x=3,y=2\text{ or }x=\frac{1}{5},y=-\frac{18}{5}$$