Solve {l}{y=frac{y_{0}}{6}(x+2)}{3x^2+4y^2-12=0}

$$\left\{ \begin{array} { l } { y = \frac { y _ { 0 } } { 6 } ( x + 2 ) } \\ { 3 x ^ { 2 } + 4 y ^ { 2 } - 12 = 0 } \end{array} \right.$$

Solve for x, y

$$x=-2\text{, }y=0$$

$$x=\frac{2\left(27-y_{0}^{2}\right)}{y_{0}^{2}+27}\text{, }y=\frac{18y_{0}}{y_{0}^{2}+27}$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Express \frac{y_{0}}{6}\left(x+2\right) as a single fraction.

$$y=\frac{y_{0}\left(x+2\right)}{6}$$

Use the distributive property to multiply y_{0} by x+2.

$$y=\frac{y_{0}x+2y_{0}}{6}$$

Subtract \frac{y_{0}x+2y_{0}}{6} from both sides.

$$y-\frac{y_{0}x+2y_{0}}{6}=0$$

Multiply both sides of the equation by 6.

$$6y-\left(y_{0}x+2y_{0}\right)=0$$

To find the opposite of y_{0}x+2y_{0}, find the opposite of each term.

$$6y-y_{0}x-2y_{0}=0$$

Add 2y_{0} to both sides. Anything plus zero gives itself.

$$6y-y_{0}x=2y_{0}$$

Consider the second equation. Add 12 to both sides. Anything plus zero gives itself.

$$3x^{2}+4y^{2}=12$$

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.

$$6y+\left(-y_{0}\right)x=2y_{0},3x^{2}+4y^{2}=12$$

Solve 6y+\left(-y_{0}\right)x=2y_{0} for y by isolating y on the left hand side of the equal sign.

$$6y+\left(-y_{0}\right)x=2y_{0}$$

Subtract \left(-y_{0}\right)x from both sides of the equation.

$$6y=y_{0}x+2y_{0}$$

Divide both sides by 6.

$$y=\frac{y_{0}}{6}x+\frac{y_{0}}{3}$$

Substitute \frac{y_{0}}{6}x+\frac{y_{0}}{3} for y in the other equation, 3x^{2}+4y^{2}=12.

$$3x^{2}+4\left(\frac{y_{0}}{6}x+\frac{y_{0}}{3}\right)^{2}=12$$

Square \frac{y_{0}}{6}x+\frac{y_{0}}{3}.

$$3x^{2}+4\left(\left(\frac{y_{0}}{6}\right)^{2}x^{2}+2\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}x+\left(\frac{y_{0}}{3}\right)^{2}\right)=12$$

Multiply 4 times \left(\frac{y_{0}}{6}\right)^{2}x^{2}+2\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}x+\left(\frac{y_{0}}{3}\right)^{2}.

$$3x^{2}+4\times \left(\frac{y_{0}}{6}\right)^{2}x^{2}+8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}x+4\times \left(\frac{y_{0}}{3}\right)^{2}=12$$

Add 3x^{2} to 4\times \left(\frac{y_{0}}{6}\right)^{2}x^{2}.

$$\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)x^{2}+8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}x+4\times \left(\frac{y_{0}}{3}\right)^{2}=12$$

Subtract 12 from both sides of the equation.

$$\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)x^{2}+8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}x+4\times \left(\frac{y_{0}}{3}\right)^{2}-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3+4\times \left(\frac{y_{0}}{6}\right)^{2} for a, 4\times 2\times \frac{y_{0}}{6}\times \frac{y_{0}}{3} for b, and \frac{4y_{0}^{2}}{9}-12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}±\sqrt{\left(8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}\right)^{2}-4\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)\left(\frac{4y_{0}^{2}}{9}-12\right)}}{2\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)}$$

Square 4\times 2\times \frac{y_{0}}{6}\times \frac{y_{0}}{3}.

$$x=\frac{-8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}±\sqrt{\frac{16y_{0}^{4}}{81}-4\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)\left(\frac{4y_{0}^{2}}{9}-12\right)}}{2\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)}$$

Multiply -4 times 3+4\times \left(\frac{y_{0}}{6}\right)^{2}.

$$x=\frac{-8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}±\sqrt{\frac{16y_{0}^{4}}{81}+\left(-\frac{4y_{0}^{2}}{9}-12\right)\left(\frac{4y_{0}^{2}}{9}-12\right)}}{2\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)}$$

Multiply -12-\frac{4y_{0}^{2}}{9} times \frac{4y_{0}^{2}}{9}-12.

$$x=\frac{-8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}±\sqrt{\frac{16y_{0}^{4}}{81}-\frac{16y_{0}^{4}}{81}+144}}{2\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)}$$

Add \frac{16y_{0}^{4}}{81} to -\frac{16y_{0}^{4}}{81}+144.

$$x=\frac{-8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}±\sqrt{144}}{2\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)}$$

Take the square root of 144.

$$x=\frac{-8\times \frac{y_{0}}{3}\times \frac{y_{0}}{6}±12}{2\left(4\times \left(\frac{y_{0}}{6}\right)^{2}+3\right)}$$

Multiply 2 times 3+4\times \left(\frac{y_{0}}{6}\right)^{2}.

$$x=\frac{-\frac{4y_{0}^{2}}{9}±12}{\frac{2y_{0}^{2}}{9}+6}$$

Now solve the equation x=\frac{-\frac{4y_{0}^{2}}{9}±12}{\frac{2y_{0}^{2}}{9}+6} when ± is plus. Add -\frac{4y_{0}^{2}}{9} to 12.

$$x=\frac{-\frac{4y_{0}^{2}}{9}+12}{\frac{2y_{0}^{2}}{9}+6}$$

Divide -\frac{4y_{0}^{2}}{9}+12 by 6+\frac{2y_{0}^{2}}{9}.

$$x=\frac{2\left(27-y_{0}^{2}\right)}{y_{0}^{2}+27}$$

Now solve the equation x=\frac{-\frac{4y_{0}^{2}}{9}±12}{\frac{2y_{0}^{2}}{9}+6} when ± is minus. Subtract 12 from -\frac{4y_{0}^{2}}{9}.

$$x=\frac{-\frac{4y_{0}^{2}}{9}-12}{\frac{2y_{0}^{2}}{9}+6}$$

Divide -\frac{4y_{0}^{2}}{9}-12 by 6+\frac{2y_{0}^{2}}{9}.

$$x=-2$$

There are two solutions for x: \frac{2\left(27-y_{0}^{2}\right)}{27+y_{0}^{2}} and -2. Substitute \frac{2\left(27-y_{0}^{2}\right)}{27+y_{0}^{2}} for x in the equation y=\frac{y_{0}}{6}x+\frac{y_{0}}{3} to find the corresponding solution for y that satisfies both equations.

$$y=\frac{y_{0}}{6}\times \frac{2\left(27-y_{0}^{2}\right)}{y_{0}^{2}+27}+\frac{y_{0}}{3}$$

Multiply \frac{y_{0}}{6} times \frac{2\left(27-y_{0}^{2}\right)}{27+y_{0}^{2}}.

$$y=\frac{2\left(27-y_{0}^{2}\right)}{y_{0}^{2}+27}\times \frac{y_{0}}{6}+\frac{y_{0}}{3}$$

Now substitute -2 for x in the equation y=\frac{y_{0}}{6}x+\frac{y_{0}}{3} and solve to find the corresponding solution for y that satisfies both equations.

$$y=\frac{y_{0}}{6}\left(-2\right)+\frac{y_{0}}{3}$$

Multiply \frac{y_{0}}{6} times -2.

$$y=-2\times \frac{y_{0}}{6}+\frac{y_{0}}{3}$$

Add -2\times \frac{y_{0}}{6} to \frac{y_{0}}{3}.

$$y=\frac{y_{0}}{3}-2\times \frac{y_{0}}{6}$$

The system is now solved.

$$y=\frac{2\left(27-y_{0}^{2}\right)}{y_{0}^{2}+27}\times \frac{y_{0}}{6}+\frac{y_{0}}{3},x=\frac{2\left(27-y_{0}^{2}\right)}{y_{0}^{2}+27}\text{ or }y=\frac{y_{0}}{3}-2\times \frac{y_{0}}{6},x=-2$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}\\x=-2\text{, }y=0\text{, }&\text{unconditionally}\\x=-\frac{2\left(y_{0}^{2}-27\right)}{y_{0}^{2}+27}\text{, }y=\frac{18y_{0}}{y_{0}^{2}+27}\text{, }&y_{0}\neq -3\sqrt{3}i\text{ and }y_{0}\neq 3\sqrt{3}i\end{matrix}\right.$$