Solve {l}{x^2+y^2=9}{x+2y=4}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 9 } \\ { x + 2 y = 4 } \end{array} \right.$$

Solve for x, y

$$x=\frac{2\sqrt{29}+4}{5}\approx 2.954065923\text{, }y=\frac{8-\sqrt{29}}{5}\approx 0.522967039$$

$$x=\frac{4-2\sqrt{29}}{5}\approx -1.354065923\text{, }y=\frac{\sqrt{29}+8}{5}\approx 2.677032961$$

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.

$$x+2y=4,y^{2}+x^{2}=9$$

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}=9.

$$y^{2}+\left(-2y+4\right)^{2}=9$$

Square -2y+4.

$$y^{2}+4y^{2}-16y+16=9$$

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

$$5y^{2}-16y+16=9$$

Subtract 9 from both sides of the equation.

$$5y^{2}-16y+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-2\right)^{2} for a, 1\times 4\left(-2\right)\times 2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 7}}{2\times 5}$$

Square 1\times 4\left(-2\right)\times 2.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 7}}{2\times 5}$$

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

$$y=\frac{-\left(-16\right)±\sqrt{256-20\times 7}}{2\times 5}$$

Multiply -20 times 7.

$$y=\frac{-\left(-16\right)±\sqrt{256-140}}{2\times 5}$$

Add 256 to -140.

$$y=\frac{-\left(-16\right)±\sqrt{116}}{2\times 5}$$

Take the square root of 116.

$$y=\frac{-\left(-16\right)±2\sqrt{29}}{2\times 5}$$

The opposite of 1\times 4\left(-2\right)\times 2 is 16.

$$y=\frac{16±2\sqrt{29}}{2\times 5}$$

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

$$y=\frac{16±2\sqrt{29}}{10}$$

Now solve the equation y=\frac{16±2\sqrt{29}}{10} when ± is plus. Add 16 to 2\sqrt{29}.

$$y=\frac{2\sqrt{29}+16}{10}$$

Divide 16+2\sqrt{29} by 10.

$$y=\frac{\sqrt{29}+8}{5}$$

Now solve the equation y=\frac{16±2\sqrt{29}}{10} when ± is minus. Subtract 2\sqrt{29} from 16.

$$y=\frac{16-2\sqrt{29}}{10}$$

Divide 16-2\sqrt{29} by 10.

$$y=\frac{8-\sqrt{29}}{5}$$

There are two solutions for y: \frac{8+\sqrt{29}}{5} and \frac{8-\sqrt{29}}{5}. Substitute \frac{8+\sqrt{29}}{5} for y in the equation x=-2y+4 to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{\sqrt{29}+8}{5}+4$$

Now substitute \frac{8-\sqrt{29}}{5} for y in the equation x=-2y+4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-2\times \frac{8-\sqrt{29}}{5}+4$$

The system is now solved.

$$x=-2\times \frac{\sqrt{29}+8}{5}+4,y=\frac{\sqrt{29}+8}{5}\text{ or }x=-2\times \frac{8-\sqrt{29}}{5}+4,y=\frac{8-\sqrt{29}}{5}$$