Solve {l}{x^2+y^2=4}{x-2y=1}

$$\left. \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 4 } \\ { x - 2 y = 1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{1-2\sqrt{19}}{5}\approx -1.543559577\text{, }y=\frac{-\sqrt{19}-2}{5}\approx -1.271779789$$

$$x=\frac{2\sqrt{19}+1}{5}\approx 1.943559577\text{, }y=\frac{\sqrt{19}-2}{5}\approx 0.471779789$$

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.

$$x-2y=1,y^{2}+x^{2}=4$$

Solve x-2y=1 for x by isolating x on the left hand side of the equal sign.

$$x-2y=1$$

Subtract -2y from both sides of the equation.

$$x=2y+1$$

Substitute 2y+1 for x in the other equation, y^{2}+x^{2}=4.

$$y^{2}+\left(2y+1\right)^{2}=4$$

Square 2y+1.

$$y^{2}+4y^{2}+4y+1=4$$

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

$$5y^{2}+4y+1=4$$

Subtract 4 from both sides of the equation.

$$5y^{2}+4y-3=0$$

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

$$y=\frac{-4±\sqrt{4^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

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

$$y=\frac{-4±\sqrt{16-4\times 5\left(-3\right)}}{2\times 5}$$

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

$$y=\frac{-4±\sqrt{16-20\left(-3\right)}}{2\times 5}$$

Multiply -20 times -3.

$$y=\frac{-4±\sqrt{16+60}}{2\times 5}$$

Add 16 to 60.

$$y=\frac{-4±\sqrt{76}}{2\times 5}$$

Take the square root of 76.

$$y=\frac{-4±2\sqrt{19}}{2\times 5}$$

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

$$y=\frac{-4±2\sqrt{19}}{10}$$

Now solve the equation y=\frac{-4±2\sqrt{19}}{10} when ± is plus. Add -4 to 2\sqrt{19}.

$$y=\frac{2\sqrt{19}-4}{10}$$

Divide -4+2\sqrt{19} by 10.

$$y=\frac{\sqrt{19}-2}{5}$$

Now solve the equation y=\frac{-4±2\sqrt{19}}{10} when ± is minus. Subtract 2\sqrt{19} from -4.

$$y=\frac{-2\sqrt{19}-4}{10}$$

Divide -4-2\sqrt{19} by 10.

$$y=\frac{-\sqrt{19}-2}{5}$$

Both solutions for y are the same: \frac{-2+\sqrt{19}}{5}. Substitute \frac{-2+\sqrt{19}}{5} for y in the equation x=2y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\times \frac{\sqrt{19}-2}{5}+1$$

Now substitute \frac{-2-\sqrt{19}}{5} for y in the equation x=2y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\times \frac{-\sqrt{19}-2}{5}+1$$

The system is now solved.

$$x=2\times \frac{\sqrt{19}-2}{5}+1,y=\frac{\sqrt{19}-2}{5}\text{ or }x=2\times \frac{-\sqrt{19}-2}{5}+1,y=\frac{-\sqrt{19}-2}{5}$$