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yes

And that is all?

yes, so when you do that what will you get?

a=a?

\[\frac{ab}{b}=\frac{c}{b}\]
\[a=\frac{c}{b}\]

Oh! now i get it, can you set up another example so i can try?

Sure ax+y=c, solve for x

Is this correct?

Really close but we need to either get rid of the y first before divide or also divide the y by a

YES! Thanks for teaching me :)

aw :(

No, first start by subtracting y from both sides

Alright, so ax+y=c
ax=c-y
x/a=c/a-y
x=c/a-y

remeber that when you divide the a you need to do it to all the terms

Where did you get z?

Oh right.. so x=c-y/a

\[x=\frac{c-y}{a}\]

zbay is correct

Inopeki got the answer i just put it in the equation maker for ease of the viewers

Yay! :D

inopeki you need to write x=(c-y)/a just to make it clearer. i see it now good work

Thanks for the help :) Is there any way you can make it a little harder to give me a challenge?

yea i will come up with one for you

Thanks.

\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\]
Solve for T_2

umm.. ill try

Nobody mentioned division by zero ?!!

Good observation @ Foolformath

I'll lay it out by steps for you

NOO. wait. Its

Cross multiplication

P1V1/T1*P2V2=T2

lcd = T_1T_2
\[T_2P_1V_1=V_2P_2T_1\]
\[T_2=\frac{V_2P_2T_1}{P_1V_1}\]

No?

What is an lcd? What makes T1 jump up?

lcd= lowest common denominator

What is that?

Is a denominator the 1 in T1?

Why do we get T1 on both sides of the line?

because you are multiplying by t1t2 but on each side one of them will cancell

Ah

Good job that was a hard one

Thanks, when am i going to get to learn this in school?

what grade are you?

8th

Oh, cool. What should i do now?

PV=nrt solve for n

And then try this \[V=\frac{4}{3}\pi r^3\] Solve for r

Divide both sides by rt, getting PV/rt=n, am i right? By the way, why is PV upper case?

Zed, isnt that for volúmes of spheres or something? exept r should be ^3

But was i right?

Zed, im going to do yours now.

yes you were right and that was the ideal gas law by the way

Oh, that was a ^3! Divide both sides with (pi)r^2 getting V/(pi)r=4/3r, am i right sofar?

What does the law state?

okay so first step is to get r^3 by itself. what do you get?

http://en.wikipedia.org/wiki/Ideal_gas_law

Wait. I take back my first answer. First you divide both sides by 4/3(pi) to isolate r^3

good. have you done powers before and their inverses?

I think so, what do you mean?

okay so we have \[r^3=\frac{3V}{4}\]
how do I get rid of the cubed part?

We should be able to just divide r by r 2 times.

3* and by itself i mean the root.

What? Can you explain that?

Why does it become 1/2?

So if r^2 becomes 1/2, then r^3 should become 1/3, right?

Correct

I screwed that up right...

What?

Sorry we want to raise both sides by the power of (1/3)

r^1/3=3V^1/3/4?

so the left hand side will be \[(r^3)^{\frac{1}{3}}=r^{\frac{3}{3}}=r^1=r\]

I dont understand.

okay the right side is good. we have \[(\frac{3V}{4})^\frac{1}{3}\]

Oh! So its r^3*1/3?

So the answer is r^3*1/3=3V^3/4

yes and that simplifies to r=(3V/4)^1/3

Oh!

Well I'm off to sleep. I'll be on tomorrow if you want more examples or help :)

Thanks for everything!