[nick198929]

Fellow Mathematicians,

I've got a question on rearranging formulae which is on changing the subject of a formula.

1. Make a the subject of the formula: A = 1/2(a + b) h

I was wondering what is the method of solving this equation? All responses on this have been a little vague and i was wondering could someone please provide a coherent explanation.

Many thanks

[HerrWarum]

_____________________________
in small steps:



times both sides by 2.

giving:



divide both sides by h:



lastly minus b from both sides:

[boy]

You wanna get a one side and everything else on the other ( means the same as expressing a in terms of A, h, and b).

So shift everything one by one

Multiply both sides by 2, divide both sides by h, take away b from both sides. a is left on its own

[Rocious]

Whatever you do to one side, you must do to the other so that they're still equal.

You don't want a fraction on the right, so multiply it by 2.

Then multiply the left by 2.

2A = h(a + b)

You don't want h on the right, so divide it...

and so on until you've just a on the right.

[nick198929]

Thanks alot! Thanks HerrWarum, Boy and Rocious. I really appreciate it, seriously. Many thanks!

[nick198929]

Is it ok if i actually put down a = 2A - b/ h ?

[boy]

No because 2A is dividing h not b or 2A - b

[HerrWarum]

Originally Posted by nick198929

Is it ok if i actually put down a = 2A - b/ h ?

a=2A - b/ h?
or
a=(2A - b)/h?

doesn't matter,
no, they're both wrong.

[nick198929]

Ok. Erm if you mind me asking why is it 2A/ h - b and not 2a-b/h ? I think i may have positioned it wrong.

[HerrWarum]

think: BIDMAS.

http://www.mathagonyaunt.co.uk/INTER.../equation.html

[boy]

Because when you divided both sides by h, you divided 2A by h on the LHS, not 2A - b

[nick198929]

Thanks alot. Is it because of the brackets? Why are the brackets transfered from (a+b) to (A/h)?

[HerrWarum]

Originally Posted by nick198929

Thanks alot. Is it because of the brackets? Why are the brackets transfered from (a+b) to (A/h)?

the brackets around: (a+b) indicate that it can be multiplied/divided to other terms.

e.g: (x+1)(x+1)

i mentioned bidmas since it's a fundemental on manipulating these types of questions which get a bit difficult later on and even involve indices e.g

i hope you do realise that:



is the same as saying:



now is it easier for you to rearrange?

[nick198929]

HerrWarrum thanks alot. I seriously don't know what my problem is regarding PEMDAS etc. For example, Make p the subject of the formula: r = 4(p+3)/ 7.

I got the answer of p = 7r - 3/ 4. I don't know what i'm doing. I think i felt that the three is a part of the r side and should be a negative.

[nick198929]

Could someone help me with my last post? Many thanks

[rnd]

Originally Posted by nick198929

r = 4(p+3)/ 7.

I got the answer

p = 7r - 3/ 4.

It's wrong I'm afraid.

What did you do? Step by step.

[nick198929]

Hey Rnd. I`ll tell you what i did step by step. First i`ll type the formula:

Make p the subject: r = 4(p + 3)/7

I first used the rules of linear equations to multiply the brackets thus giving:

r = 4p + 12/ 7

I then turned the equation around so that the current subject is on the right hand side:

4p + 12/7 = r

Then i started with the 7 which is divide thus being converted into multiplication thus being infront of the r and then i looked at the 12. The 4is being multiplied by p so it's a divide. That must be coverted into a negative so i now have:

p = 7r -12/ 4

[rnd]

OK. You tried.

Here's what I would do.



Multiply by 7



Divide by 4



Subtract 3



Write it the other way around if you like..

[nick198929]

Intriguing ok so you don't actually multiply the brackets. Thanks Rnd. But again i seem to be doing something wrong. I got the answer p = 7r -3/ 4.

[rnd]

Originally Posted by nick198929

Thanks Rnd.

You're welcome.

Originally Posted by nick198929

I got the answer p = 7r -3/ 4.

How?