Solve
`(a) 3/sqrtx+4/sqrtx=sqrtx`
`(b) (y^3-(7)^(7/2))/sqrty=0`

I. 7/sqrtx + 5/sqrtx = sqrtx II. y^(2) -((12)^(5//2))/sqrty = 0

y = sqrtx + 1/sqrtx

1/sqrtx (sqrtx+1/sqrtx)^2

(4)/(sqrtx) + (7)/(sqrtx) = sqrtx, y^(2)-((11)^(5/2))/(sqrty) = 0

y = x+sqrtx+1/sqrtx

In the following questions two equations numbered I and II are given. You have to solve both the equations "I. "6/sqrtx+(16)/sqrtx=sqrtx" II. "y^5-((11xx2)^(1//2))/sqrty=0

2/sqrtx + 3/sqrty = 2

2/sqrtx + 3/sqrty = 2

2/sqrtx + 3/sqrty = 2

y = sqrt(2 -sqrtx)