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 February 22nd, 2017, 04:52 PM #1 Newbie   Joined: Feb 2017 From: USA Posts: 28 Thanks: 1 Assistance with simplifying an equation Hello all, I would like to know where I am going wrong with this equation. S = 2πrh + 2πr^2 solve for h I took it step by step and thought I did it correctly. Can someone point me to a concise resource on the algebraic rules for simplifying equations, especially fractions with variables? This is what I did: S = 2πrh + 2πr^2 - 2πr^2 S- 2πr^2 = 2πrh S- 2πr^2 = 2πrh -------------------- 2πr h = S- 2πr^2 / 2πr I then made the mistake of trying to reduce the 2πr term into the numerator. Why is this not allowed? Thanks!!
February 22nd, 2017, 05:13 PM   #2
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 Originally Posted by chopnhack Hello all, I would like to know where I am going wrong with this equation. S = 2πrh + 2πr^2 solve for h I took it step by step and thought I did it correctly. Can someone point me to a concise resource on the algebraic rules for simplifying equations, especially fractions with variables? This is what I did: S = 2πrh + 2πr^2 - 2πr^2 S- 2πr^2 = 2πrh S- 2πr^2 = 2πrh -------------------- 2πr h = S- 2πr^2 / 2πr I then made the mistake of trying to reduce the 2πr term into the numerator. Why is this not allowed? Thanks!!
Just a shot in the dark here, but what you wrote: h = S- 2πr^2 / 2πr is not correct. You used no parenthesis. What the formula is:
$\displaystyle h = \frac{S - 2 \pi r^2}{2 \pi r}$

In order to do the division S would have to be divisible by $\displaystyle 2 \pi r$ and it is not.

-Dan

 February 22nd, 2017, 05:20 PM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 The rules are exactly the same as what you learned in first year algebra. First, get all terms involving the variable of interest on one side of the equation and all other terms on the other side. $S = 2 \pi rh + 2 \pi r^2 \implies 2 \pi rh = S - 2 \pi r^2.$ Now reduce the one side to just the variable of interest. $\dfrac{1}{2 \pi r} * 2 \pi r h = \dfrac{1}{2 \pi r} * (S - 2 \pi r^2) \implies h = \dfrac{S}{2 \pi r} - r.$ EDIT: My answer and topsquark's are equivalent. Thanks from chopnhack Last edited by JeffM1; February 22nd, 2017 at 05:22 PM.
 February 22nd, 2017, 07:28 PM #4 Newbie   Joined: Feb 2017 From: USA Posts: 28 Thanks: 1 Thanks guys! I see where I went wrong. Thanks Jeff, that was the direction I was taking, just made the mistake of not paying attention to the simplification of the correct term.

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