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

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

Solve for x, y

$$x=\frac{1-\sqrt{7}}{2}\approx -0.822875656\text{, }y=\frac{-\sqrt{7}-1}{4}\approx -0.911437828$$

$$x=\frac{\sqrt{7}+1}{2}\approx 1.822875656\text{, }y=\frac{\sqrt{7}-1}{4}\approx 0.411437828$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 4.

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

Consider the second equation. Subtract 2y from both sides.

$$x-2y=1$$

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.

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

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, 4y^{2}+x^{2}=4.

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

Square 2y+1.

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

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

$$8y^{2}+4y+1=4$$

Subtract 4 from both sides of the equation.

$$8y^{2}+4y-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4+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}.

$$y=\frac{-4±\sqrt{4^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

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

$$y=\frac{-4±\sqrt{16-4\times 8\left(-3\right)}}{2\times 8}$$

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

$$y=\frac{-4±\sqrt{16-32\left(-3\right)}}{2\times 8}$$

Multiply -32 times -3.

$$y=\frac{-4±\sqrt{16+96}}{2\times 8}$$

Add 16 to 96.

$$y=\frac{-4±\sqrt{112}}{2\times 8}$$

Take the square root of 112.

$$y=\frac{-4±4\sqrt{7}}{2\times 8}$$

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

$$y=\frac{-4±4\sqrt{7}}{16}$$

Now solve the equation y=\frac{-4±4\sqrt{7}}{16} when ± is plus. Add -4 to 4\sqrt{7}.

$$y=\frac{4\sqrt{7}-4}{16}$$

Divide -4+4\sqrt{7} by 16.

$$y=\frac{\sqrt{7}-1}{4}$$

Now solve the equation y=\frac{-4±4\sqrt{7}}{16} when ± is minus. Subtract 4\sqrt{7} from -4.

$$y=\frac{-4\sqrt{7}-4}{16}$$

Divide -4-4\sqrt{7} by 16.

$$y=\frac{-\sqrt{7}-1}{4}$$

Both solutions for y are the same: \frac{-1+\sqrt{7}}{4}. Substitute \frac{-1+\sqrt{7}}{4} for y in the equation x=2y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\times \frac{\sqrt{7}-1}{4}+1$$

Now substitute \frac{-1-\sqrt{7}}{4} for y in the equation x=2y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\times \frac{-\sqrt{7}-1}{4}+1$$

The system is now solved.

$$x=2\times \frac{\sqrt{7}-1}{4}+1,y=\frac{\sqrt{7}-1}{4}\text{ or }x=2\times \frac{-\sqrt{7}-1}{4}+1,y=\frac{-\sqrt{7}-1}{4}$$