The Student Room Group
Reply 1
Although 1 and 2 are a slightly different type to 3, the general idea is:
- get all terms containing the letter you want to factorise to one side
- factorise
- You can then usually divide the equation by the coefficient of h.

For example:
Make t the subject of
tc=tu+bc
tc-tu=bc
t(c-u)=bc
Now we divide both sides by (c-u) to get:
t=bc/(c-u)

I would suggest you attempt your questions, and if you get stuck, post your answer/how far you have got, and it is easier to see where you are going wrong/right! :smile:
Reply 2
As rec53 said, you need to add/subtract/multiply terms to get all the terms containing h on one side of the equation, then factorise to get h outside a bracket.

I'll show you how to do the first one, and then you can hopefully do the other two.

2RT=mgh+K2h\large \frac{2}{RT} = mgh+\normalsize{K}^2 h
2RT=h(mg+K2)\large \frac{2}{RT} = h(mg+\normalsize{K}^2) (factorising the right-hand side so that h is outside the bracket)
h=2RT(mg+K2)\large h = \frac{2}{RT(mg+K^2)} (getting h on one side of the equation)

:smile:
Reply 3
thanks guys, I get how fatuous_philomath got that answer but I still cant work out the second one.....

can anyone help?
thanks
Reply 4
Someone else may be willing to just give the answer to this, but it would be better if you post your attempt, even if you are certain it is wrong, and we can try and explain where you are going wrong.

(you just need to work through the method that I & fatuous_philomath explained)
Reply 5
richard321


2) 2lh=6x^2+2xh

Thanks


2lh=6x^2+2xh (get all of the h values on one side, -2xh)
2lh-2xh=6x^2 (now you can factorise the h values)
h(2l-2x)=6x^2 (and devide by the part you don't want, /(2l-2x) )

That should be correct
Reply 6
iknowmole
what did you get for the first? I think you need logs.

K^2h is meant to be read as (K^2)h=h*K^2 I believe, in which case, we don't need logs for any questions.

(the original post I was replying to has now disappeared, left for clarification)
Reply 7
yeah sorry i'm just used to reading formulas with superscript instead of ^.

Forget what i said about logs.
Reply 8
right, thanks alot guys I know understand this, i think...

for the 3rd one I got:

h= the square root of 10lp/y

is that right?
Reply 9
richard321
right, thanks alot guys I know understand this, i think...

for the 3rd one I got:

h= the square root of 10lp/y

is that right?

yep :smile:
Reply 10
richard321

Any help on these and guides on how to do them will be much appreciated.
Thanks


the best way to approch these in my opinion is to get all the values of h on one side, (what ever you do to one side you do to the other, e.g. if you have as=b+y and you want to get y on the other side you take y from both sides).

From hear you need to remove any parts which are unwanted (what ever you do to one side you do to the other), you may need to factorise first, e.g. yt-3t=t(y-3).

Then you need to get the single letter to the right power. for example if you have t^3=hy-65f you can cube root each side to get the answer.
Reply 11
nice one guys, thanks for the help , much appreciated.