Question 1927
i am getting tired now lol. Here you go. They look right (although the final one looks wierd to me)

1. {{{(w-3)^(1/2) = (4w+15)^(1/2)}}}...square both sides
{{{(w-3) = (4w+15)}}}
3w=-18
w=-6

check...put -6 into the original, and the 2 sides should still equal each other.

2. {{{(x^2-5x+2)^1/2=x}}}... square both sides
(x^2-5x+2)=x^2

therefore -5x+2=0
5x=2
x=2/5

Again check by putting this answer into the original..both sides should be the same.

3. {{{(5/(y-3)) = 1 + ((y+7)/(2y-6))}}}
{{{((2*5)/(2(y-3))) = 1 + ((y+7)/(2y-6))}}}..i have just scaled the first term by 2 so i can add it to the other term --> both will have the same denominator now...

{{{(10/(2y-6)) = 1 + ((y+7)/(2y-6))}}}

{{{(10/(2y-6)) - ((y+7)/(2y-6)) = 1}}}

{{{(10 - (y+7))/(2y-6) = 1}}}
{{{(10-y-7)/(2y-6) = 1}}}
{{{(3-y)=(2y-6)}}}
3y=9
y=3

Now, put this into the equations and the denominators are zero...divide by zero is NOT healthy, so this is te only worry i have, but it is nearly 1am and i am tired and i cannot think straight.

cheers
Jon.