SOLUTION: OK, I am so not worthy of your math wisdom :) (thank you for helping). There are still a few I am lost on....I hope you have the time. :) I am supposed to solve the equation and

Question 1927: OK, I am so not worthy of your math wisdom :) (thank you for helping).
There are still a few I am lost on....I hope you have the time. :)
I am supposed to solve the equation and check for extraneous solutions:
(w-3)^1/2=(4w+15)^1/2
AND(x^2-5x+2)^1/2=x
OR
5/(y-3)=1+[(y+7)/(2y-6)]
Jon, you are helping me so much, I really do appreciate it cuz I am so lost on this stuff. Tutoring like this is my last hope. :)
Answer by longjonsilver(2297) (Show Source): You can put this solution on YOUR website!
i am getting tired now lol. Here you go. They look right (although the final one looks wierd to me)
1. ...square both sides

3w=-18
w=-6
check...put -6 into the original, and the 2 sides should still equal each other.
2. ... 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.
..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...

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.
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