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

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

Solve for x, y

$$x=-\frac{2\sqrt{6}}{3}\approx -1.632993162\text{, }y=-\frac{\sqrt{3}}{3}\approx -0.577350269$$

$$x=\frac{2\sqrt{6}}{3}\approx 1.632993162\text{, }y=\frac{\sqrt{3}}{3}\approx 0.577350269$$

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. Express \frac{\sqrt{2}}{4}x as a single fraction.

$$y=\frac{\sqrt{2}x}{4}$$

Subtract \frac{\sqrt{2}x}{4} from both sides.

$$y-\frac{\sqrt{2}x}{4}=0$$

Multiply both sides of the equation by 4.

$$4y-\sqrt{2}x=0$$

Reorder the terms.

$$-\sqrt{2}x+4y=0$$

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.

$$\left(-\sqrt{2}\right)x+4y=0,4y^{2}+x^{2}=4$$

Solve \left(-\sqrt{2}\right)x+4y=0 for x by isolating x on the left hand side of the equal sign.

$$\left(-\sqrt{2}\right)x+4y=0$$

Subtract 4y from both sides of the equation.

$$\left(-\sqrt{2}\right)x=-4y$$

Divide both sides by -\sqrt{2}.

$$x=2\sqrt{2}y$$

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

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

Square 2\sqrt{2}y.

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

Add 4y^{2} to \left(2\sqrt{2}\right)^{2}y^{2}.

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

Subtract 4 from both sides of the equation.

$$\left(\left(2\sqrt{2}\right)^{2}+4\right)y^{2}-4=0$$

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

$$y=\frac{0±\sqrt{0^{2}-4\left(\left(2\sqrt{2}\right)^{2}+4\right)\left(-4\right)}}{2\left(\left(2\sqrt{2}\right)^{2}+4\right)}$$

Square 1\times 0\times 2\times 2\sqrt{2}.

$$y=\frac{0±\sqrt{-4\left(\left(2\sqrt{2}\right)^{2}+4\right)\left(-4\right)}}{2\left(\left(2\sqrt{2}\right)^{2}+4\right)}$$

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

$$y=\frac{0±\sqrt{-48\left(-4\right)}}{2\left(\left(2\sqrt{2}\right)^{2}+4\right)}$$

Multiply -48 times -4.

$$y=\frac{0±\sqrt{192}}{2\left(\left(2\sqrt{2}\right)^{2}+4\right)}$$

Take the square root of 192.

$$y=\frac{0±8\sqrt{3}}{2\left(\left(2\sqrt{2}\right)^{2}+4\right)}$$

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

$$y=\frac{0±8\sqrt{3}}{24}$$

Now solve the equation y=\frac{0±8\sqrt{3}}{24} when ± is plus.

$$y=\frac{\sqrt{3}}{3}$$

Now solve the equation y=\frac{0±8\sqrt{3}}{24} when ± is minus.

$$y=-\frac{\sqrt{3}}{3}$$

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

$$x=2\sqrt{2}\times \frac{\sqrt{3}}{3}$$

Now substitute -\frac{\sqrt{3}}{3} for y in the equation x=2\sqrt{2}y and solve to find the corresponding solution for x that satisfies both equations.

$$x=2\sqrt{2}\left(-\frac{\sqrt{3}}{3}\right)$$

The system is now solved.

$$x=2\sqrt{2}\times \frac{\sqrt{3}}{3},y=\frac{\sqrt{3}}{3}\text{ or }x=2\sqrt{2}\left(-\frac{\sqrt{3}}{3}\right),y=-\frac{\sqrt{3}}{3}$$