Solve the system: $\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{3}}=2$, $(\sqrt{2}-1)x+(\sqrt{3}-1)y=5-\sqrt{2}-\sqrt{3}$

Question

Solve the system: $\begin{array}{|l} \dfrac{x}{\sqrt{2}}+\dfrac{y}{\sqrt{3}}=2 \\ (\sqrt{2}-1)x+(\sqrt{3}-1)y=5-\sqrt{2}-\sqrt{3} \end{array}$

So, I started by multiplying the first equation by $\sqrt{2}\cdot\sqrt{3}=\sqrt{6}$: $\begin{array}{|l} \sqrt{3}x+\sqrt{2}y=2\sqrt{6} \\ (\sqrt{2}-1)x+(\sqrt{3}-1)y=5-\sqrt{2}-\sqrt{3} \end{array}$

How can I solve from here?

Answer 1

Using Cramer's method: $$|A|=\begin{vmatrix}\frac1{\sqrt2}&\frac1{\sqrt3}\\ \sqrt2-1&\sqrt3-1\end{vmatrix}=\frac{\sqrt3-1}{\sqrt2}-\frac{\sqrt2-1}{\sqrt3}=\frac{1+\sqrt2-\sqrt3}{\sqrt6}\\ |A_x|=\begin{vmatrix}2&\frac1{\sqrt3}\\ 5-\sqrt2-\sqrt3&\sqrt3-1\end{vmatrix}=2(\sqrt3-1)-\frac{5-\sqrt2-\sqrt3}{\sqrt3}=\frac{1+\sqrt2-\sqrt3}{\sqrt3}\\ x=\frac{|A_x|}{|A|}=\sqrt2$$ Can you find $y$?

Answer 2

If you're comfortable with linear algebra, you have a $2\times 2$ matrix which is invertible. If not, try solving for $x$ (or $y$) in the first equation and plug it into the second equation