Solve the following system of equation: [((1), sqrt(2)x+sqrt(3)y=0),((2), x+y=sqrt(3)-sqrt(2))]?

1 Answer
Apr 19, 2016

$\left\{\begin{matrix}x = \frac{3 \sqrt{2} - 2 \sqrt{3}}{\sqrt{6} - 2} \\ y = \frac{\sqrt{6} - 2}{\sqrt{2} - \sqrt{3}}\end{matrix}\right.$

Explanation:

From $\left(1\right)$ we have

$\sqrt{2} x + \sqrt{3} y = 0$

Dividing both sides by $\sqrt{2}$ gives us

$x + \frac{\sqrt{3}}{\sqrt{2}} y = 0 \text{ (*)}$

If we subtract $\text{(*)}$ from $\left(2\right)$ we obtain

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

$\implies \left(1 - \frac{\sqrt{3}}{\sqrt{2}}\right) y = \sqrt{3} - \sqrt{2}$

$\implies y = \frac{\sqrt{3} - \sqrt{2}}{1 - \frac{\sqrt{3}}{\sqrt{2}}} = \frac{\sqrt{6} - 2}{\sqrt{2} - \sqrt{3}}$

If we substitute the value we found for $y$ back into $\text{(*)}$ we get

$x + \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{6} - 2}{\sqrt{2} - \sqrt{3}} = 0$

$\implies x + \frac{3 \sqrt{2} - 2 \sqrt{3}}{2 - \sqrt{6}} = 0$

$\implies x = - \frac{3 \sqrt{2} - 2 \sqrt{3}}{2 - \sqrt{6}} = \frac{3 \sqrt{2} - 2 \sqrt{3}}{\sqrt{6} - 2}$

Thus, we arrive at the solution

$\left\{\begin{matrix}x = \frac{3 \sqrt{2} - 2 \sqrt{3}}{\sqrt{6} - 2} \\ y = \frac{\sqrt{6} - 2}{\sqrt{2} - \sqrt{3}}\end{matrix}\right.$