Solve {l}{x-2y=0}{x^2/a^2+y^2/b^2=1}

$$\left\{ \begin{array} { l } { x - 2 y = 0 } \\ { \frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{2ab}{\sqrt{a^{2}+4b^{2}}}\text{, }y=-\frac{ab}{\sqrt{a^{2}+4b^{2}}}$$

$$x=\frac{2ab}{\sqrt{a^{2}+4b^{2}}}\text{, }y=\frac{ab}{\sqrt{a^{2}+4b^{2}}}\text{, }a\neq 0\text{ and }b\neq 0$$

Steps Using Substitution and Quadratic Formula

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.

$$x-2y=0,\frac{1}{b^{2}}y^{2}+\frac{1}{a^{2}}x^{2}=1$$

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

$$x-2y=0$$

Subtract -2y from both sides of the equation.

$$x=2y$$

Substitute 2y for x in the other equation, \frac{1}{b^{2}}y^{2}+\frac{1}{a^{2}}x^{2}=1.

$$\frac{1}{b^{2}}y^{2}+\frac{1}{a^{2}}\times \left(2y\right)^{2}=1$$

Square 2y.

$$\frac{1}{b^{2}}y^{2}+\frac{1}{a^{2}}\times 4y^{2}=1$$

Multiply a^{-2} times 4y^{2}.

$$\frac{1}{b^{2}}y^{2}+\frac{4}{a^{2}}y^{2}+\frac{0}{a^{2}}y+\frac{0}{a^{2}}=1$$

Add \frac{1}{b^{2}}y^{2} to \frac{4}{a^{2}}y^{2}.

$$\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)y^{2}+\frac{0}{a^{2}}y+\frac{0}{a^{2}}=1$$

Subtract 1 from both sides of the equation.

$$\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)y^{2}+\frac{0}{a^{2}}y+\frac{0}{a^{2}}-1=0$$

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

$$y=\frac{-\frac{0}{a^{2}}±\sqrt{\left(\frac{0}{a^{2}}\right)^{2}-4\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)\left(-1\right)}}{2\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)}$$

Square a^{-2}\times 0\times 2\times 2.

$$y=\frac{-\frac{0}{a^{2}}±\sqrt{-4\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)\left(-1\right)}}{2\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)}$$

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

$$y=\frac{-\frac{0}{a^{2}}±\sqrt{\left(-\frac{4}{b^{2}}-\frac{16}{a^{2}}\right)\left(-1\right)}}{2\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)}$$

Multiply -\frac{4}{b^{2}}-\frac{16}{a^{2}} times -1.

$$y=\frac{-\frac{0}{a^{2}}±\sqrt{\frac{4}{b^{2}}+\frac{16}{a^{2}}}}{2\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)}$$

Take the square root of \frac{4}{b^{2}}+\frac{16}{a^{2}}.

$$y=\frac{-\frac{0}{a^{2}}±\frac{2\sqrt{a^{2}+4b^{2}}}{|a||b|}}{2\left(\frac{1}{b^{2}}+\frac{4}{a^{2}}\right)}$$

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

$$y=\frac{0±\frac{2\sqrt{a^{2}+4b^{2}}}{|a||b|}}{\frac{2}{b^{2}}+\frac{8}{a^{2}}}$$

Now solve the equation y=\frac{0±\frac{2\sqrt{a^{2}+4b^{2}}}{|a||b|}}{\frac{2}{b^{2}}+\frac{8}{a^{2}}} when ± is plus.

$$y=\frac{a^{2}b^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}}$$

Now solve the equation y=\frac{0±\frac{2\sqrt{a^{2}+4b^{2}}}{|a||b|}}{\frac{2}{b^{2}}+\frac{8}{a^{2}}} when ± is minus.

$$y=-\frac{a^{2}b^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}}$$

There are two solutions for y: \frac{b^{2}a^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}} and -\frac{b^{2}a^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}}. Substitute \frac{b^{2}a^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}} for y in the equation x=2y to find the corresponding solution for x that satisfies both equations.

$$x=2\times \frac{a^{2}b^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}}$$

Now substitute -\frac{b^{2}a^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}} for y in the equation x=2y and solve to find the corresponding solution for x that satisfies both equations.

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

The system is now solved.

$$x=2\times \frac{a^{2}b^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}},y=\frac{a^{2}b^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}}\text{ or }x=2\left(-\frac{a^{2}b^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}}\right),y=-\frac{a^{2}b^{2}}{|a||b|\sqrt{a^{2}+4b^{2}}}$$

Solve for x, y (complex solution)

$$x=2\left(a^{2}+4b^{2}\right)^{-\frac{1}{2}}ab\text{, }y=\left(a^{2}+4b^{2}\right)^{-\frac{1}{2}}ab$$

$$x=-2\left(a^{2}+4b^{2}\right)^{-\frac{1}{2}}ab\text{, }y=-\left(a^{2}+4b^{2}\right)^{-\frac{1}{2}}ab\text{, }a\neq -2ib\text{ and }a\neq 2ib\text{ and }a\neq 0\text{ and }b\neq 0$$