# Simultaneous equations

• Jul 6th 2008, 02:00 PM
hciR
Simultaneous equations
I'm trying to solve these as opposed to using the matrix method (which I don't fully understand yet) but can't see where I go wrong. e2 should be 1.8 but I'm getting 1.61. Could the difference be due to not using enough decimal places or something I'm not doing right?

http://www.fileden.com/files/2007/8/...n_P1000406.JPG
• Jul 6th 2008, 10:24 PM
earboth
Quote:

Originally Posted by hciR
I'm trying to solve these as opposed to using the matrix method (which I don't fully understand yet) but can't see where I go wrong. e2 should be 1.8 but I'm getting 1.61. Could the difference be due to not using enough decimal places or something I'm not doing right?

You have used the wrong numbers. For instance:

$\frac3{8200} = 0.00036585... = 3.6585 \cdot 10^{-4}~\neq~3.6585^{-4}$
• Jul 7th 2008, 03:51 AM
hciR
I realise i've wrote it down wrong, but thats just because I was too lazy to write x10^-4. When putting it in my calculator I wrote it in correctly (3.659times10x^y-4)
• Jul 7th 2008, 01:06 PM
meymathis
First issue is that it is hard to read the image. So it is difficult to check your arithmetic. Based on:
$
( \frac{1}{8200}+\frac{1}{1500}+\frac{1}{3900} ) e_1 - \frac{1}{1500}e_2 = \frac{3}{8200}
$

$
-\frac{1}{1500}e_1 + ( \frac{1}{3900} + \frac{1}{1500} + \frac{1}{8200} ) e_2= \frac{3}{3900}
$

I got your answers of 1.38 and 1.61 (the latter really should be rounded to 1.62 since I am getting 1.617). My guess is a) either the starting equations are wrong, or b) the answer of 1.8 is wrong.

Somewhat beside the point but why are you adding (1)+(2)? This is unnecessary. Just take your (3) (which is equivalent to (1) and substitute it into (2).

Incidentally, the matrix method is exactly the same as working with the equations, only it it more compact (less notation).

Finally, if you want to be lazy but still be correct, you just need to add E:
$3.65 \cdot 10^{-4}$ becomes 3.65E-4
• Jul 7th 2008, 03:29 PM
hciR
Ok you got the same answers which is good. I'm not quite sure why I added (1) and (2) but I can see that you don't need to now.
I'm following a course, the PDF is here with the equations on page 16 and the solution on page 19.
Would the "E" button just be labelled "E" or do you have to press shift and something else?
Thanks
• Jul 7th 2008, 08:35 PM
meymathis
Quote:

Originally Posted by hciR
Ok you got the same answers which is good. I'm not quite sure why I added (1) and (2) but I can see that you don't need to now.
I'm following a course, the PDF is here with the equations on page 16 and the solution on page 19.

It looks to me that the equations that both you and I have written down match each other and match what is on page 17 (which is equivalent to what is on 16). Take your solutions and plug them into the original equations. If they both work, then you got it right and they have a typo. If not then we're doing something silly.

Quote:

Originally Posted by hciR
Would the "E" button just be labelled "E" or do you have to press shift and something else?
Thanks

I just meant when posting a message or writing down on paper. But yes you can do it on a calculator, and on mine (some obscure and now out-dated TI-85), I think you just hit EE button. (i.e. 3.65EE-4 does what you want).

See your calculator manual for details.(Nerd)
• Jul 8th 2008, 04:24 AM
hciR
I put the answers back into the equation and it does work so maybe there is a typo, I will go through what they have on the pdf and see if I can spot something.
I found if you press shift then log, it gives you a 10^x symbo l(Cool)