Question 17.

In such questions, we should be trying to complete the squares. 

We see a $xy$ term; we need to accommodate that in a square that has both x and y terms. 

Since there is only one other term with x, we also need to have it entirely in the square. 

$\left(2x-y\right)^{^2}=4x^2+y^2-4xy$

Using this in the given equation, we are left with $\left(2x-y\right)^{^2}+3y^2+3-6y$

This can be written as $\left(2x-y\right)^{^2}+3\left(y^2+1-2y\right)$

$\left(2x-y\right)^{^2}+3\left(y-1\right)^2=0$

Since both the squares add up to 0, this is only possible when the squares themselves are 0

This would give us y=1 from the second term, and using that, we get x= 1/2 from the first term. 

Therefore the value of 4x+5y will be 2+5 = 7

Hence, 7 is the correct answer.