Question

If $x,y$ are rational numbers such that $(x+2y)+(x-3y)\sqrt{6}=(x-y-2)\sqrt{5}+(2x+y-2)$, then

• A. $x+y=6$
• B. $x+y=5$
• C. $x+y=4$
• D. $x+y=3$

Answer 1

The correct option is C $x+y=4$

$(x+2y)+(x-3y)\sqrt{6}=(x-y-2)\sqrt{5}+(2x+y-2)$
Since $x,y\in Q$, so coefficient of $\sqrt{6}$ and $\sqrt{5}$ should be zero
Since $x,y\in Q$
$\therefore x+2y=2x+y-2$ .........(i)
$x-y-2=0$ .........(ii)
$x-3y=0$ .........(iii)
On solving (ii) and (iii) $x=3$ , $y=1$
$\therefore x+y=4$
These values also satisfy the original equation