ab=c
Solve for a
Do i need to divide both sides by b?

- anonymous

ab=c
Solve for a
Do i need to divide both sides by b?

- chestercat

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- anonymous

yes

- anonymous

And that is all?

- anonymous

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

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## More answers

- anonymous

a=a?

- anonymous

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

- anonymous

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

- anonymous

Sure ax+y=c, solve for x

- anonymous

Ok, divide ax with a, then divide c with a getting ax/a+y=c/a.
You then simplify that getting x+y=c/a.
Then you take both sides minus y getting x=c/a-y

- anonymous

Is this correct?

- anonymous

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

- anonymous

YES! Thanks for teaching me :)

- anonymous

aw :(

- anonymous

No, first start by subtracting y from both sides

- anonymous

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

- anonymous

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

- anonymous

Where did you get z?

- anonymous

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

- anonymous

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

- anonymous

zbay is correct

- anonymous

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

- anonymous

Yay! :D

- anonymous

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

- anonymous

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

- anonymous

yea i will come up with one for you

- anonymous

Thanks.

- anonymous

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

- anonymous

umm.. ill try

- anonymous

Nobody mentioned division by zero ?!!

- anonymous

Good observation @ Foolformath

- anonymous

Well to begin with i need to multiply both sides with P2V2, getting P1V1/T1*P2V2, right?\[P _{1}V _{1}divT _{1}+P _{2}V _{2}=T2\]

- anonymous

I'll lay it out by steps for you

- anonymous

NOO. wait. Its

- anonymous

Cross multiplication

- anonymous

P1V1/T1*P2V2=T2

- anonymous

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

- anonymous

Oh right, i see it now. You wanted me to put the things i was multiplying by(V2P2) in by the other things that were multiplied. That would mean my answer is right, but yours is simplified.

- anonymous

No?

- anonymous

No because you need to multiply both sides by the lcd of \[T_1T_2\] first then we cancel one of the varibles out because they are the same on top and bottom. then we can break T_2 apart from the rest of the left side by division, your answer would look close but in the end it would be wrong

- anonymous

What is an lcd? What makes T1 jump up?

- anonymous

lcd= lowest common denominator

- anonymous

What is that?

- anonymous

Is a denominator the 1 in T1?

- anonymous

The denominator is what is on the bottom of the fraction, so when we do the left side we get
\[\frac{T_1T_2P_11V_1}{T_1}\]
and because we have the same thing on the top and bottom we can cancel out the denominator leving us with
\[T_2P_1V_1\]
and we do the same operation on the right side

- anonymous

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

- anonymous

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

- anonymous

Ah

- anonymous

Good job that was a hard one

- anonymous

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

- anonymous

what grade are you?

- anonymous

8th

- anonymous

Not sure when you will cover that but it's a chemistry equation compariing similar gasses to determine preasure, temp, or volume. You might touch it in highschool sometime.

- anonymous

Oh, cool. What should i do now?

- anonymous

PV=nrt solve for n

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

good question and i don't have an answer for you but in that equation they are all upercase but the little n but i made a mistake and didn't capitilize the RT

- anonymous

But was i right?

- anonymous

Zed, im going to do yours now.

- anonymous

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

- anonymous

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

- anonymous

What does the law state?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

good. have you done powers before and their inverses?

- anonymous

I think so, what do you mean?

- anonymous

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

- anonymous

Oh, i have done powers but i dont recall doing inverses, is that like squareroot and cubicroot? I guess we need to get the "cubicalroot" out of r^3.

- anonymous

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

- anonymous

3* and by itself i mean the root.

- anonymous

not quite see let's try this example to help explain it
\[x^2=4\]\[x=\sqrt{4}\]\[x=4^{\frac{1}{2}}\]\[x=2\]

- anonymous

What? Can you explain that?

- anonymous

okay so when we bring powers (eg 3) over the equation sign they turn to 1/powers (eg 1/3).
if we look at squared when we bring it over it turns to 1/2 which is also known as the square root.

- anonymous

Why does it become 1/2?

- anonymous

the may be abit hard to follow but this is why.
Let f(x)=y
\[y=x^7\]
\[\ln y=7 \ln x\]
\[lnx=(lny)/7\]
\[x=e^{(lny)/7}\]
\[Inverse = f^{-1}(x)=e^{(lnx)/7}\]
\[ f^{-1}(x)=e^{lnx^\frac{1}{7}}\]\[=x^\frac{1}{7}\]

- anonymous

Now usually when learning this we just accept this as fact that \[x^n=x^{\frac{1}{n}}\] where n is a positive integer

- anonymous

That was probably way too much information at once just use the rule in the above post and try to solve \[r^3=\frac{3V}{4}\]

- anonymous

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

- anonymous

Correct

- anonymous

That means that i have to maake r^3 to 1/3 and convert 1/3 to 4/12(to merge it with the right side). i then convert the right sides fractions too, getting 9V/12. Now i should be able to merge them, right? so i get r=13/12?

- anonymous

I screwed that up right...

- anonymous

What?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

I dont understand.

- anonymous

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

- anonymous

remember on the left hand side we had r^3 and when raised it to the power of 1/3 the powers need to be multiplied.

- anonymous

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

- anonymous

yes exactly it's one of the rules for powers.
here are others if you wanted to learn and practice http://www.math.com/school/subject2/lessons/S2U2L2DP.html

- anonymous

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

- anonymous

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

- anonymous

Oh!

- anonymous

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

- anonymous

Thanks for everything!

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