The answer is A..........
X=2h, y=3k
Substitute these values into equations.
y+2x = 4 ------> 3k+2*2h=4 -----> 3k +4h =4
2/y - 3/2x = 1-----> 2/3k -3/(2*2h) = 1 ------> 2/3k - 3/4h =1
We have a system of equations now.
3k +4h =4 ------> 3k = 4-4h ( Substitute 3k in the 2nd equation.)
2/3k - 3/4h =1
2/(4-4h) -3/4h = 1
2/(2(2-2h)) - 3/4h = 1
1/(2-2h) -3/4h - 1=0
4h/4h(2-2h) -3(2-2h)/4h(2-2h) - 4h(2-2h)/4h(2-2h) =0
(4h- 3(2-2h) - 4h(2-2h))/4h(2-2h) = 0
Numerator should be = 0
4h- 3(2-2h) - 4h(2-2h)=0
Denominator cannot be = 0
4h(2-2h)≠0
Solve equation for numerator=0
4h- 3(2-2h) - 4h(2-2h)=0
4h - 6+6h-8h+8h² =0
8h² +2h -6=0
4h² + h-3 =0
(4h-3)(h+1)=0
4h-3=0, h+1=0
h=3/4 or h=-1
Check which
4h(2-2h)≠0
1) h= 3/4 , 4*3/4(2-2*3/4)=3*(2-6)= -12 ≠0, so we can use h= 3/4
2)h=-1, 4(-1)(2-2*(-1)) =-4*4=-16 ≠0, so we can use h= -1, also.
h=3/4, then 3k = 4-4*3/4 =4 - 3=1 , 3k =1, k=1/3
h=-1, then 3k = 4-4*(-1) =8 , 3k=8, k=8/3
So,
if h=3/4, then k=1/3,
and if h=-1, then k=8/3 .
The smallest natural number is 1. The largest natural number does not exists because there will always be one even larger. But if you are asking the number of all natural numbers or the size of natural number set then 
is your answer.
Answer:
Since the base is the same, we can just multiply the exponents, and when we do so we get y^6.
