We know that (a+b)2=a2+b2+2ab
take a=y, and b=1/y
a2+b2=(a+b)2-2ab
y2+1/y2=(y+1/y)2-2x y x 1/y
y2+1/y2 =(y+1/y)2-2..........................(1)
7(y+1/y)-2(y2+1/y2)=9
7(y+1/y)-2[(y+1/y)2-2]=9..............from equation(1)
taking (y+1/y)=z we get
7z-2[z2-2]=9
=> 7z-2z2+4=9
=> -2z2+7z-5=0................(2)
solving equation (2) we get
2z2-7z+5=0
=> z(2z-5)-1(2z-5)=0
=> (z-1)(2z-5)=0
=>(z-1)=0 or (2z-5)=0
z=1 or, z=5/2
Now,
y+1/y=1 or y+1/y=5/2
=> y2+1=y
=> y2-y+1=0 [this is not possible]
Now,
y+1/y=5/2
=>y2+1=5y/2
=> 2y2+2=5y
=> 2y2-5y+2=0
=>2y(y-2)-1(y-2)=0
=> (2y-1)(y-2)=0
=>2y-1=0 ,y-2=0
number of integral solution of the equation
y=1/2, y=2