Solve {l}{y=-1/2kx}{x^2+2y^2=4}

$$\left. \begin{array} { l } { y = - \frac { 1 } { 2 k } x } \\ { x ^ { 2 } + 2 y ^ { 2 } = 4 } \end{array} \right.$$

Solve for x, y

$$x=2\sqrt{\frac{2}{2k^{2}+1}}k\text{, }y=-\sqrt{\frac{2}{2k^{2}+1}}$$

$$x=-2\sqrt{\frac{2}{2k^{2}+1}}k\text{, }y=\sqrt{\frac{2}{2k^{2}+1}}\text{, }k\neq 0$$

Steps Using Substitution and Quadratic Formula

Consider the first equation. Multiply both sides of the equation by 2k.

$$y\times 2k=\left(-\frac{1}{2k}\right)x\times 2k$$

Express \left(-\frac{1}{2k}\right)x as a single fraction.

$$y\times 2k=\frac{-x}{2k}\times 2k$$

Express \frac{-x}{2k}\times 2 as a single fraction.

$$y\times 2k=\frac{-x\times 2}{2k}k$$

Cancel out 2 in both numerator and denominator.

$$y\times 2k=\frac{-x}{k}k$$

Express \frac{-x}{k}k as a single fraction.

$$y\times 2k=\frac{-xk}{k}$$

Cancel out k in both numerator and denominator.

$$y\times 2k=-x$$

Add x to both sides.

$$y\times 2k+x=0$$

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.

$$2ky+x=0,x^{2}+2y^{2}=4$$

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

$$2ky+x=0$$

Subtract x from both sides of the equation.

$$2ky=-x$$

Divide both sides by 2k.

$$y=\left(-\frac{1}{2k}\right)x$$

Substitute \left(-\frac{1}{2k}\right)x for y in the other equation, x^{2}+2y^{2}=4.

$$x^{2}+2\left(\left(-\frac{1}{2k}\right)x\right)^{2}=4$$

Square \left(-\frac{1}{2k}\right)x.

$$x^{2}+2\left(-\frac{1}{2k}\right)^{2}x^{2}=4$$

Add x^{2} to 2\left(-\frac{1}{2k}\right)^{2}x^{2}.

$$\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)x^{2}=4$$

Subtract 4 from both sides of the equation.

$$\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)x^{2}-4=0$$

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

$$x=\frac{0±\sqrt{0^{2}-4\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)\left(-4\right)}}{2\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)}$$

Square 2\times 0\times 2\left(-\frac{1}{2k}\right).

$$x=\frac{0±\sqrt{-4\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)\left(-4\right)}}{2\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)}$$

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

$$x=\frac{0±\sqrt{\left(-4-\frac{2}{k^{2}}\right)\left(-4\right)}}{2\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)}$$

Multiply -4-\frac{2}{k^{2}} times -4.

$$x=\frac{0±\sqrt{16+\frac{8}{k^{2}}}}{2\left(1+2\left(-\frac{1}{2k}\right)^{2}\right)}$$

Take the square root of 16+\frac{8}{k^{2}}.

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

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

$$x=\frac{0±\frac{2\sqrt{4k^{2}+2}}{|k|}}{2+\frac{1}{k^{2}}}$$

Now solve the equation x=\frac{0±\frac{2\sqrt{4k^{2}+2}}{|k|}}{2+\frac{1}{k^{2}}} when ± is plus.

$$x=\frac{4k^{2}}{|k|\sqrt{4k^{2}+2}}$$

Now solve the equation x=\frac{0±\frac{2\sqrt{4k^{2}+2}}{|k|}}{2+\frac{1}{k^{2}}} when ± is minus.

$$x=-\frac{4k^{2}}{|k|\sqrt{4k^{2}+2}}$$

There are two solutions for x: \frac{4k^{2}}{\sqrt{4k^{2}+2}|k|} and -\frac{4k^{2}}{\sqrt{4k^{2}+2}|k|}. Substitute \frac{4k^{2}}{\sqrt{4k^{2}+2}|k|} for x in the equation y=\left(-\frac{1}{2k}\right)x to find the corresponding solution for y that satisfies both equations.

$$y=\left(-\frac{1}{2k}\right)\times \frac{4k^{2}}{|k|\sqrt{4k^{2}+2}}$$

Now substitute -\frac{4k^{2}}{\sqrt{4k^{2}+2}|k|} for x in the equation y=\left(-\frac{1}{2k}\right)x and solve to find the corresponding solution for y that satisfies both equations.

$$y=\left(-\frac{1}{2k}\right)\left(-\frac{4k^{2}}{|k|\sqrt{4k^{2}+2}}\right)$$

The system is now solved.

$$y=\left(-\frac{1}{2k}\right)\times \frac{4k^{2}}{|k|\sqrt{4k^{2}+2}},x=\frac{4k^{2}}{|k|\sqrt{4k^{2}+2}}\text{ or }y=\left(-\frac{1}{2k}\right)\left(-\frac{4k^{2}}{|k|\sqrt{4k^{2}+2}}\right),x=-\frac{4k^{2}}{|k|\sqrt{4k^{2}+2}}$$

Solve for x, y (complex solution)

$$x=-2\sqrt{2}\left(2k^{2}+1\right)^{-\frac{1}{2}}k\text{, }y=\sqrt{2}\left(2k^{2}+1\right)^{-\frac{1}{2}}$$

$$x=2\sqrt{2}\left(2k^{2}+1\right)^{-\frac{1}{2}}k\text{, }y=-\sqrt{2}\left(2k^{2}+1\right)^{-\frac{1}{2}}\text{, }k\neq -\frac{\sqrt{2}i}{2}\text{ and }k\neq \frac{\sqrt{2}i}{2}\text{ and }k\neq 0$$