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

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

Solve for x, y

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

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

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.

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

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

$$x-2y=4$$

Subtract -2y from both sides of the equation.

$$x=2y+4$$

Substitute 2y+4 for x in the other equation, y^{2}+x^{2}=13.

$$y^{2}+\left(2y+4\right)^{2}=13$$

Square 2y+4.

$$y^{2}+4y^{2}+16y+16=13$$

Add y^{2} to 4y^{2}.

$$5y^{2}+16y+16=13$$

Subtract 13 from both sides of the equation.

$$5y^{2}+16y+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 2^{2} for a, 1\times 4\times 2\times 2 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-16±\sqrt{16^{2}-4\times 5\times 3}}{2\times 5}$$

Square 1\times 4\times 2\times 2.

$$y=\frac{-16±\sqrt{256-4\times 5\times 3}}{2\times 5}$$

Multiply -4 times 1+1\times 2^{2}.

$$y=\frac{-16±\sqrt{256-20\times 3}}{2\times 5}$$

Multiply -20 times 3.

$$y=\frac{-16±\sqrt{256-60}}{2\times 5}$$

Add 256 to -60.

$$y=\frac{-16±\sqrt{196}}{2\times 5}$$

Take the square root of 196.

$$y=\frac{-16±14}{2\times 5}$$

Multiply 2 times 1+1\times 2^{2}.

$$y=\frac{-16±14}{10}$$

Now solve the equation y=\frac{-16±14}{10} when ± is plus. Add -16 to 14.

$$y=-\frac{2}{10}$$

Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.

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

Now solve the equation y=\frac{-16±14}{10} when ± is minus. Subtract 14 from -16.

$$y=-\frac{30}{10}$$

Divide -30 by 10.

$$y=-3$$

There are two solutions for y: -\frac{1}{5} and -3. Substitute -\frac{1}{5} for y in the equation x=2y+4 to find the corresponding solution for x that satisfies both equations.

$$x=2\left(-\frac{1}{5}\right)+4$$

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

$$x=-\frac{2}{5}+4$$

Add -\frac{1}{5}\times 2 to 4.

$$x=\frac{18}{5}$$

Now substitute -3 for y in the equation x=2y+4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\left(-3\right)+4$$

Multiply 2 times -3.

$$x=-6+4$$

Add -3\times 2 to 4.

$$x=-2$$

The system is now solved.

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