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

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And that is all?
yes, so when you do that what will you get?

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Other answers:

\[\frac{ab}{b}=\frac{c}{b}\] \[a=\frac{c}{b}\]
Oh! now i get it, can you set up another example so i can try?
Sure ax+y=c, solve for x
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
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
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
\[\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\] Solve for T_2
umm.. ill try
Nobody mentioned division by zero ?!!
Good observation @ Foolformath
Well to begin with i need to multiply both sides with P2V2, getting P1V1/T1*P2V2, right?\[P _{1}V _{1}divT _{1}+P _{2}V _{2}=T2\]
I'll lay it out by steps for you
NOO. wait. Its
Cross multiplication
lcd = T_1T_2 \[T_2P_1V_1=V_2P_2T_1\] \[T_2=\frac{V_2P_2T_1}{P_1V_1}\]
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.
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
What is an lcd? What makes T1 jump up?
lcd= lowest common denominator
What is that?
Is a denominator the 1 in T1?
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
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
Good job that was a hard one
Thanks, when am i going to get to learn this in school?
what grade are you?
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.
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
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
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?
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?
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.
We should be able to just divide r by r 2 times.
3* and by itself i mean the root.
not quite see let's try this example to help explain it \[x^2=4\]\[x=\sqrt{4}\]\[x=4^{\frac{1}{2}}\]\[x=2\]
What? Can you explain that?
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.
Why does it become 1/2?
the may be abit hard to follow but this is why. Let f(x)=y \[y=x^7\] \[\ln y=7 \ln x\] \[lnx=(lny)/7\] \[x=e^{(lny)/7}\] \[Inverse = f^{-1}(x)=e^{(lnx)/7}\] \[ f^{-1}(x)=e^{lnx^\frac{1}{7}}\]\[=x^\frac{1}{7}\]
Now usually when learning this we just accept this as fact that \[x^n=x^{\frac{1}{n}}\] where n is a positive integer
That was probably way too much information at once just use the rule in the above post and try to solve \[r^3=\frac{3V}{4}\]
So if r^2 becomes 1/2, then r^3 should become 1/3, right?
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?
I screwed that up right...
Sorry we want to raise both sides by the power of (1/3)
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 \[(\frac{3V}{4})^\frac{1}{3}\]
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.
Oh! So its r^3*1/3?
yes exactly it's one of the rules for powers. here are others if you wanted to learn and practice
So the answer is r^3*1/3=3V^3/4
yes and that simplifies to r=(3V/4)^1/3
Well I'm off to sleep. I'll be on tomorrow if you want more examples or help :)
Thanks for everything!

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