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What do you think is a good first step?

umm maybe..\[\sqrt{2}\times \sqrt{1/y}=10\]

If the problem
\[ 2y^{-1/2} - 10 \]
or
\[ (2y)^{-1/2} = 10 \]

*Is

\[2y ^{-1/2}=10\]

How did you get 5?

oh wait never mind. You just divided by 2

C'mon. Look at the expression before hand. I divided both sides by 2.

then we should turn y^(-1/2) into \[\sqrt{1/y}\] right?

Ok.
\[\sqrt{1/y}=5\] I square both sides right? so 1/y =25

yes and hence y = ...

25?

If 1/y = 25, then it can NOT be the case y = 25.

I'm not sure

If 1/x = 2, then x = 1/2. If 1/z = 1/3, then z = 3

Hence if
1/y = 25
then
y = ...

I don't understand how to find y. I kind of forgot how to do the reciprocal thing .Can u explain

Scary.
\[ 2 \times \frac{1}{2} = 1 \]
Agreed?

two times one half equals one, yes?

yes

Suppose now that 2x = 1. What does x equal?

1/2

but i mean in the form of 1/y=25. I don't understand this one

If 1/y = a, then y = 1/a

If 1/y = 25, then y = ... ?

ummm

I thought I would just times y/1 to both sides

If 1/y = a, then y = 1/a

Hence if 1/y = 25, then y = ....

1/y=25
times by y/1 to both sides... 25y???

yes

Now suppose 1/y = 25, then
\[ 25 \times y = 1 \]
That is
\[ 25y = 1 \]
Agreed so far?

ok. 25y=1
Oh then divide by 25 so I get 1/25

thank you!