Solve {l}{y=4/15x}{x^2+4y^2=4}

$$\left\{ \begin{array} { l } { y = \frac { 4 } { 15 } x } \\ { x ^ { 2 } + 4 y ^ { 2 } = 4 } \end{array} \right.$$

Solve for y, x

$$x=-\frac{30}{17}\approx -1.764705882\text{, }y=-\frac{8}{17}\approx -0.470588235$$

$$x=\frac{30}{17}\approx 1.764705882\text{, }y=\frac{8}{17}\approx 0.470588235$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract \frac{4}{15}x from both sides.

$$y-\frac{4}{15}x=0$$

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.

$$y-\frac{4}{15}x=0,x^{2}+4y^{2}=4$$

Solve y-\frac{4}{15}x=0 for y by isolating y on the left hand side of the equal sign.

$$y-\frac{4}{15}x=0$$

Subtract -\frac{4}{15}x from both sides of the equation.

$$y=\frac{4}{15}x$$

Substitute \frac{4}{15}x for y in the other equation, x^{2}+4y^{2}=4.

$$x^{2}+4\times \left(\frac{4}{15}x\right)^{2}=4$$

Square \frac{4}{15}x.

$$x^{2}+4\times \frac{16}{225}x^{2}=4$$

Multiply 4 times \frac{16}{225}x^{2}.

$$x^{2}+\frac{64}{225}x^{2}=4$$

Add x^{2} to \frac{64}{225}x^{2}.

$$\frac{289}{225}x^{2}=4$$

Subtract 4 from both sides of the equation.

$$\frac{289}{225}x^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+4\times \left(\frac{4}{15}\right)^{2} for a, 4\times 0\times \frac{4}{15}\times 2 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{289}{225}\left(-4\right)}}{2\times \frac{289}{225}}$$

Square 4\times 0\times \frac{4}{15}\times 2.

$$x=\frac{0±\sqrt{-4\times \frac{289}{225}\left(-4\right)}}{2\times \frac{289}{225}}$$

Multiply -4 times 1+4\times \left(\frac{4}{15}\right)^{2}.

$$x=\frac{0±\sqrt{-\frac{1156}{225}\left(-4\right)}}{2\times \frac{289}{225}}$$

Multiply -\frac{1156}{225} times -4.

$$x=\frac{0±\sqrt{\frac{4624}{225}}}{2\times \frac{289}{225}}$$

Take the square root of \frac{4624}{225}.

$$x=\frac{0±\frac{68}{15}}{2\times \frac{289}{225}}$$

Multiply 2 times 1+4\times \left(\frac{4}{15}\right)^{2}.

$$x=\frac{0±\frac{68}{15}}{\frac{578}{225}}$$

Now solve the equation x=\frac{0±\frac{68}{15}}{\frac{578}{225}} when ± is plus.

$$x=\frac{30}{17}$$

Now solve the equation x=\frac{0±\frac{68}{15}}{\frac{578}{225}} when ± is minus.

$$x=-\frac{30}{17}$$

There are two solutions for x: \frac{30}{17} and -\frac{30}{17}. Substitute \frac{30}{17} for x in the equation y=\frac{4}{15}x to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{15}\times \frac{30}{17}$$

Multiply \frac{4}{15} times \frac{30}{17} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{8}{17}$$

Now substitute -\frac{30}{17} for x in the equation y=\frac{4}{15}x and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{4}{15}\left(-\frac{30}{17}\right)$$

Multiply \frac{4}{15} times -\frac{30}{17} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{8}{17}$$

The system is now solved.

$$y=\frac{8}{17},x=\frac{30}{17}\text{ or }y=-\frac{8}{17},x=-\frac{30}{17}$$