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Maths help required...

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bradzy_c 07 Nov 2012
Anyone help me out with a simultanious equation?

(4.3)P1 = (-1.2)P2 + (28.1)
(4.1)P1 = (5.6)P2 + (9.6)

I need to find the value of either P1 or P2. I've tried for P2 and only seem to turn up with 8.167 which can't be right because when I use it for substitution I get two different numbers for P1 which is wrong.

Cheers,
Brad
In reply to bradzy_c:

They don't have a coherent solution (i.e. no value of P1 and P2 satisfies both equations).
bradzy_c 07 Nov 2012
In reply to victim of mathematics: They must do as I wouldn't have been assigned it otherwise. Here's how far I got:

P1 = (-1.2P2 + 28.1)/4.3
P1 = (5.6P2 + 9.6)/4.1

These are my first rearrangements

(-1.2P2 + 28.1)/4.3 = (5.6P2 + 9.6)/4.1

Their is my comparison

4.1(-1.2P2 + 28.1) = (4.3(5.6P2 + 9.6)

I then cross multiplied

115.21 - 4.92P2 = 41.28 + 24.08P2

I then expanded

24.08P2 - 4.92P2 = 41.21 + 115.21

Here is my re-arranged for a second time

19.16P2 = 156.49

Simplification

P2 = 156.49/19.16
P2 = 8.167

But for some reason when I substitute it's incorrect.
 Andy Hardy 07 Nov 2012
In reply to bradzy_c:

I got 5.81 and 2.54, not tried subbing back into the original eqns tho.

(Started by multiplying everything by 10, then adding the 2 equations together. Rearrange to get P1 in terms of P2, then put the expression for P1 back into one of the original equations)
In reply to 999thAndy: I got P1 = 5.82 and P2 = 2.55 so close to your numbers
In reply to bradzy_c:

>
> 115.21 - 4.92P2 = 41.28 + 24.08P2
>
> I then expanded
>
> 24.08P2 - 4.92P2 = 41.21 + 115.21
>

OK, I messed up plugging my answers back into the original equations. I get 5.82 and 2.55. Your mistake is in this rearrangement ^

115.21-4.92p2=41.28+24.08P2

rearranges to:

115.21-41.22=24.08P2+4.92P2
andic 07 Nov 2012
In reply to bradzy_c:

I got 2.550 and 5.8

It does have an answer plot the two eqns together: http://dl.dropbox.com/u/23995729/sim%20eqns.pdf
bradzy_c 07 Nov 2012
In reply to 999thAndy: Without being a massive pain in the arse, could you put step by step how you did it?
 SARS 07 Nov 2012
In reply to bradzy_c:

Multiple eqn 1 by 4.1 and eqn 2 by 4.3 and then subtract the two.
 James Malloch 07 Nov 2012
In reply to bradzy_c:
> (In reply to victim of mathematics) They must do as I wouldn't have been assigned it otherwise. Here's how far I got:
>
> I then cross multiplied
>
> 115.21 - 4.92P2 = 41.28 + 24.08P2
>
> I then expanded
>
> 24.08P2 - 4.92P2 = 41.21 + 115.21
>

You have gone wrong in this step. You want to put both "P2"'s on one side and the standard numbers on the other.

So you should have added 4.92P2 to both sides and subtracted 41.28 from both sides. So it would be...

115.21 - 41.21 = 24.08P2 + 4.92P2

And then divide by the coefficiant of P2 once they have been summed together.
bradzy_c 07 Nov 2012
In reply to James Malloch: Cheers guys! Appreciate everyone's help!I'll buy you all a kit-kat
In reply to andic:
> (In reply to bradzy_c)
>
> I got 2.550 and 5.8
>
> It does have an answer plot the two eqns together: http://dl.dropbox.com/u/23995729/sim%20eqns.pdf

Me too: P2 = 2.55 and P1 = 5.82

 thebrookster 07 Nov 2012
In reply to bradzy_c:

1) Re-arrange both equations to have P1 and P2 on one side, known number on the other.

(4.3)P1 + (1.2)P2 = 28.1
(4.1)P1 - (5.6)P2 = 9.6

2) multiply top equation by 5.6

(24.08)P1 + (6.72)P2 = 157.36

3) multiply bottom equation by 1.2

(4.92)P1 - (6-72)P2 = 9.6

4)add both equations together

(29.08) = 168.88 (P2 terms cancel each other out)

5) Re-arrange and solver for P1, P1 = 5.82

6) Sub P1 back into an earlier equation, say from step 2, and solve for P2

P2 = 2.55

My double check tells me this is correct.

So answer is:

P1 = 5.82
P2 = 2.55
In reply to bradzy_c:

This is where you went wrong:

24.08P2 - 4.92P2 = 41.21 + 115.21

That 24.08P2 should be -24.08P2 - you forgot to change the sign when you took it across from the other side of the equation.




 thebrookster 07 Nov 2012
In reply to thebrookster:

Incidently, using my method by re-arranging the formulas, I naturally got one positive P2 term and one negative term, hence neutralising this term (I could see this in the original question, hence this method).

If you had only positive terms, then for step 4 you would simply subtract one equation from the other. Same for only negative terms (two minus' make a positive).
 Andy Hardy 07 Nov 2012
In reply to bradzy_c:
> (In reply to 999thAndy) Without being a massive pain in the arse, could you put step by step how you did it?

If I do that you won't learn. Follow the outline I have and see what you arrive at.
(Plus I'm typing on a phone...)
bradzy_c 07 Nov 2012
In reply to bradzy_c: Yeh got my head round it all now. Basically didn't change positive to a negative and vice versa. Silly mistake! Cheers guys!

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