Solve the system: root(2) x + root(3) y = 3 root(2) and root(3) x - root(2) y = 2 root(3).

Step-by-step solution

1. 1

   Choose elimination over substitution. Solving either equation for a variable would drag radicals into the denominator immediately. Elimination is cleaner because the $y$-coefficients $\sqrt3$ and $-\sqrt2$ multiply to the same $\sqrt6$.

2. 2

   Scale the first equation by $\sqrt2$. $\sqrt2\left(\sqrt2\,x+\sqrt3\,y\right)=\sqrt2\cdot 3\sqrt2\ \Longrightarrow\ 2x+\sqrt6\,y=6.$

3. 3

   Scale the second equation by $\sqrt3$. $\sqrt3\left(\sqrt3\,x-\sqrt2\,y\right)=\sqrt3\cdot 2\sqrt3\ \Longrightarrow\ 3x-\sqrt6\,y=6.$

4. 4

   Add to eliminate $y$. The $\sqrt6\,y$ terms cancel exactly: $5x=12\ \Longrightarrow\ x=\frac{12}{5}.$

5. 5

   Back-substitute and rationalise. From the first original equation, $\sqrt3\,y=3\sqrt2-\frac{12\sqrt2}{5}=\frac{3\sqrt2}{5}\ \Longrightarrow\ y=\frac{3\sqrt2}{5\sqrt3}=\frac{3\sqrt6}{15}=\frac{\sqrt6}{5}.$

6. 6

   Check both equations numerically. With $x=2.4$ and $y=0.4898979$: the first left side is $4.2426407=3\sqrt2$, and the second is $3.4641016=2\sqrt3$. Both hold.