Solve
`sqrt(y+6)-sqrt(y-1)=sqrt(y-9)`

find the value of y: sqrt(y+6) - sqrt(y-1)=sqrt(y-9)

The equation of tangent to the curve sqrt(x)-sqrt(y)=1 at (9,4) is

The equation of normal to the curve sqrt(x)-sqrt(y)=1 at (9,4) is

(sqrt(x)+sqrt(y))^(2)=x+y+2sqrt(xy) and sqrt(x)sqrt(y)=sqrt(xy) , where x and y are positive real numbers . If x=2sqrt(5)+sqrt(3) and y=2sqrt(5)-sqrt(3) , then x^(4)+y^(4) =

Solve the following system of linear equations (with rational denominator) by using the method of elimination : sqrt(3)x - sqrt(2)y = sqrt (3) and sqrt(5)x + sqrt(3)y = sqrt(2) .

sqrt(x/y) + sqrt(y/x) = 6

Solve: (2)/(sqrt(x))-(3)/(sqrt(y))=2 and (4)/(sqrt(x))-(9)/(sqrt(y))=-1