Solve {l}{x^2/a+y^2/b=1}{y=2x+6}

$$\left. \begin{array} { l } { \frac { x ^ { 2 } } { a } + \frac { y ^ { 2 } } { b } = 1 } \\ { y = 2 x + 6 } \end{array} \right.$$

Solve for x, y

$$\left\{\begin{matrix}x=-\frac{\sqrt{ab\left(4a+b-36\right)}+12a}{4a+b}\text{, }y=\frac{2\left(-\sqrt{ab\left(4a+b-36\right)}+3b\right)}{4a+b}\text{; }x=\frac{\sqrt{ab\left(4a+b-36\right)}-12a}{4a+b}\text{, }y=\frac{2\left(\sqrt{ab\left(4a+b-36\right)}+3b\right)}{4a+b}\text{, }&\left(a\geq -\frac{b}{4}+9\text{ and }a>0\text{ and }b>0\right)\text{ or }\left(a=-\frac{b}{4}+9\text{ and }b\neq 0\text{ and }b<36\right)\text{ or }\left(a\neq -\frac{b}{4}\text{ and }a\leq -\frac{b}{4}+9\text{ and }b<0\text{ and }a>0\right)\text{ or }\left(a=-\frac{b}{4}+9\text{ and }b>0\text{ and }b\neq 36\right)\text{ or }\left(a\neq -\frac{b}{4}\text{ and }a\leq -\frac{b}{4}+9\text{ and }a<0\text{ and }b>0\right)\\x=\frac{b-36}{24}\text{, }y=\frac{b+36}{12}\text{, }&a=-\frac{b}{4}\text{ and }b\neq 0\end{matrix}\right.$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by ab, the least common multiple of a,b.

$$bx^{2}+ay^{2}=ab$$

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

$$y-2x=6$$

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.

$$y-2x=6,bx^{2}+ay^{2}=ab$$

Solve y-2x=6 for y by isolating y on the left hand side of the equal sign.

$$y-2x=6$$

Subtract -2x from both sides of the equation.

$$y=2x+6$$

Substitute 2x+6 for y in the other equation, bx^{2}+ay^{2}=ab.

$$bx^{2}+a\left(2x+6\right)^{2}=ab$$

Square 2x+6.

$$bx^{2}+a\left(4x^{2}+24x+36\right)=ab$$

Multiply a times 4x^{2}+24x+36.

$$bx^{2}+4ax^{2}+24ax+36a=ab$$

Add bx^{2} to 4ax^{2}.

$$\left(4a+b\right)x^{2}+24ax+36a=ab$$

Subtract ab from both sides of the equation.

$$\left(4a+b\right)x^{2}+24ax+36a-ab=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute b+a\times 2^{2} for a, a\times 6\times 2\times 2 for b, and a\left(36-b\right) for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24a±\sqrt{\left(24a\right)^{2}-4\left(4a+b\right)a\left(36-b\right)}}{2\left(4a+b\right)}$$

Square a\times 6\times 2\times 2.

$$x=\frac{-24a±\sqrt{576a^{2}-4\left(4a+b\right)a\left(36-b\right)}}{2\left(4a+b\right)}$$

Multiply -4 times b+a\times 2^{2}.

$$x=\frac{-24a±\sqrt{576a^{2}+\left(-16a-4b\right)a\left(36-b\right)}}{2\left(4a+b\right)}$$

Multiply -4b-16a times a\left(36-b\right).

$$x=\frac{-24a±\sqrt{576a^{2}-4a\left(36-b\right)\left(4a+b\right)}}{2\left(4a+b\right)}$$

Add 576a^{2} to -4\left(b+4a\right)a\left(36-b\right).

$$x=\frac{-24a±\sqrt{4ab\left(4a+b-36\right)}}{2\left(4a+b\right)}$$

Take the square root of 4ab\left(-36+4a+b\right).

$$x=\frac{-24a±2\sqrt{ab\left(4a+b-36\right)}}{2\left(4a+b\right)}$$

Multiply 2 times b+a\times 2^{2}.

$$x=\frac{-24a±2\sqrt{ab\left(4a+b-36\right)}}{8a+2b}$$

Now solve the equation x=\frac{-24a±2\sqrt{ab\left(4a+b-36\right)}}{8a+2b} when ± is plus. Add -24a to 2\sqrt{ab\left(-36+4a+b\right)}.

$$x=\frac{2\sqrt{ab\left(4a+b-36\right)}-24a}{8a+2b}$$

Divide -24a+2\sqrt{ab\left(-36+4a+b\right)} by 2b+8a.

$$x=\frac{\sqrt{ab\left(4a+b-36\right)}-12a}{4a+b}$$

Now solve the equation x=\frac{-24a±2\sqrt{ab\left(4a+b-36\right)}}{8a+2b} when ± is minus. Subtract 2\sqrt{ab\left(-36+4a+b\right)} from -24a.

$$x=\frac{-2\sqrt{ab\left(4a+b-36\right)}-24a}{8a+2b}$$

Divide -24a-2\sqrt{ab\left(-36+4a+b\right)} by 2b+8a.

$$x=-\frac{\sqrt{ab\left(4a+b-36\right)}+12a}{4a+b}$$

There are two solutions for x: \frac{-12a+\sqrt{ab\left(-36+4a+b\right)}}{b+4a} and -\frac{12a+\sqrt{ab\left(-36+4a+b\right)}}{b+4a}. Substitute \frac{-12a+\sqrt{ab\left(-36+4a+b\right)}}{b+4a} for x in the equation y=2x+6 to find the corresponding solution for y that satisfies both equations.

$$y=2\times \frac{\sqrt{ab\left(4a+b-36\right)}-12a}{4a+b}+6$$

Now substitute -\frac{12a+\sqrt{ab\left(-36+4a+b\right)}}{b+4a} for x in the equation y=2x+6 and solve to find the corresponding solution for y that satisfies both equations.

$$y=2\left(-\frac{\sqrt{ab\left(4a+b-36\right)}+12a}{4a+b}\right)+6$$

The system is now solved.

$$y=2\times \frac{\sqrt{ab\left(4a+b-36\right)}-12a}{4a+b}+6,x=\frac{\sqrt{ab\left(4a+b-36\right)}-12a}{4a+b}\text{ or }y=2\left(-\frac{\sqrt{ab\left(4a+b-36\right)}+12a}{4a+b}\right)+6,x=-\frac{\sqrt{ab\left(4a+b-36\right)}+12a}{4a+b}$$

Solve for x, y (complex solution)

$$\left\{\begin{matrix}x=-\frac{\sqrt{ab\left(4a+b-36\right)}+12a}{4a+b}\text{, }y=\frac{2\left(-\sqrt{ab\left(4a+b-36\right)}+3b\right)}{4a+b}\text{; }x=\frac{\sqrt{ab\left(4a+b-36\right)}-12a}{4a+b}\text{, }y=\frac{2\left(\sqrt{ab\left(4a+b-36\right)}+3b\right)}{4a+b}\text{, }&a\neq -\frac{b}{4}\text{ and }a\neq 0\text{ and }b\neq 0\\x=\frac{b-36}{24}\text{, }y=\frac{b+36}{12}\text{, }&a=-\frac{b}{4}\text{ and }b\neq 0\end{matrix}\right.$$