Solve {l}{2y=15x}{x^2+y^2=229/9}

$$\left. \begin{array} { l } { 2 y = 15 x } \\ { x ^ { 2 } + y ^ { 2 } = \frac { 229 } { 9 } } \end{array} \right.$$

Solve for y, x

$$x=-\frac{2}{3}\approx -0.666666667\text{, }y=-5$$

$$x=\frac{2}{3}\approx 0.666666667\text{, }y=5$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Subtract 15x from both sides.

$$2y-15x=0$$

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.

$$2y-15x=0,x^{2}+y^{2}=\frac{229}{9}$$

Solve 2y-15x=0 for y by isolating y on the left hand side of the equal sign.

$$2y-15x=0$$

Subtract -15x from both sides of the equation.

$$2y=15x$$

Divide both sides by 2.

$$y=\frac{15}{2}x$$

Substitute \frac{15}{2}x for y in the other equation, x^{2}+y^{2}=\frac{229}{9}.

$$x^{2}+\left(\frac{15}{2}x\right)^{2}=\frac{229}{9}$$

Square \frac{15}{2}x.

$$x^{2}+\frac{225}{4}x^{2}=\frac{229}{9}$$

Add x^{2} to \frac{225}{4}x^{2}.

$$\frac{229}{4}x^{2}=\frac{229}{9}$$

Subtract \frac{229}{9} from both sides of the equation.

$$\frac{229}{4}x^{2}-\frac{229}{9}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times \left(\frac{15}{2}\right)^{2} for a, 1\times 0\times 2\times \frac{15}{2} for b, and -\frac{229}{9} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times \frac{229}{4}\left(-\frac{229}{9}\right)}}{2\times \frac{229}{4}}$$

Square 1\times 0\times 2\times \frac{15}{2}.

$$x=\frac{0±\sqrt{-4\times \frac{229}{4}\left(-\frac{229}{9}\right)}}{2\times \frac{229}{4}}$$

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

$$x=\frac{0±\sqrt{-229\left(-\frac{229}{9}\right)}}{2\times \frac{229}{4}}$$

Multiply -229 times -\frac{229}{9}.

$$x=\frac{0±\sqrt{\frac{52441}{9}}}{2\times \frac{229}{4}}$$

Take the square root of \frac{52441}{9}.

$$x=\frac{0±\frac{229}{3}}{2\times \frac{229}{4}}$$

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

$$x=\frac{0±\frac{229}{3}}{\frac{229}{2}}$$

Now solve the equation x=\frac{0±\frac{229}{3}}{\frac{229}{2}} when ± is plus.

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

Now solve the equation x=\frac{0±\frac{229}{3}}{\frac{229}{2}} when ± is minus.

$$x=-\frac{2}{3}$$

There are two solutions for x: \frac{2}{3} and -\frac{2}{3}. Substitute \frac{2}{3} for x in the equation y=\frac{15}{2}x to find the corresponding solution for y that satisfies both equations.

$$y=\frac{15}{2}\times \frac{2}{3}$$

Multiply \frac{15}{2} times \frac{2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=5$$

Now substitute -\frac{2}{3} for x in the equation y=\frac{15}{2}x and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{15}{2}\left(-\frac{2}{3}\right)$$

Multiply \frac{15}{2} times -\frac{2}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-5$$

The system is now solved.

$$y=5,x=\frac{2}{3}\text{ or }y=-5,x=-\frac{2}{3}$$