Question

Hello,
This is the latest that has stumped me... hang on I'm trying to see how to upload a little pic of the equation....

Answer 1

Press attach file below

Answer 2

by "make \(L_1\) the subject" it is asking to solve for \(L_1\) correct?

Answer 3

Cool thanks, I couldn't see how to post it straight up, but I see it now and have uploaded it :) thanks

Answer 4

correct

Answer 5

\[\large \frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2+L_3}\]

Answer 6

L1? It looks like L3

Answer 7

yes thats right - what program do you use to be able to do that please??

Answer 8

Ahh, Sorry. \(L_3\)*

Answer 9

program??...We can simply solve for it.

Answer 10

First we need to find a common denominator between \(L~,~ L_1~,~ \text{and}~ L_2+L_3\).

Answer 11

haha no i mean to be able to format the maths equation. not to solve the actual problem, jsut to be able to type in the equation like you have (rather than like what i had to do - go and scan my text book! ) There's gotta be a better way - looks like you know how to type it in using some program... you see what i was getting at?

Answer 12

Ah, this is just LaTex formatting. I simply typed in `\[\frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2+L_3}\]`

Answer 13

I've got to go right now, hopefully someone else can finish helping you with your question. Good luck!

Answer 14

ok back to maths. So, to me the lowest common denominator would be L * L1 * L2 * L3 (im trying to say L times L1 times L2 times L3). Is that correct?

Answer 15

ok thanks for the hint.

Answer 16

Multiply each fraction by the least common denominator of all three fractions.

Answer 17

\(LCD = LL_1(L_2 + L_3)\)
Notice the plus sign between L2 and L3.

Answer 18

Im trying it again with your advice now, thank you!

Answer 19

\(\large LL_1(L_2 + L_3) \dfrac{1}{L}=LL_1(L_2 + L_3) \dfrac{1}{L_1}+LL_1(L_2 + L_3) \dfrac{1}{L_2+L_3}\)

Answer 20

The purpose of multiplying each fraction by the LCD is to then eliminate all denominators.

Answer 21

\( \cancel{L}L_1(L_2 + L_3) \dfrac{1}{\cancel{L}}=L\cancel{L_1}(L_2 + L_3) \dfrac{1}{\cancel{L_1}}+LL_1(\cancel{L_2 + L_3}) \dfrac{1}{\cancel{L_2+L_3}}\)

\( L_1(L_2 + L_3) =L(L_2 + L_3)+LL_1\)

Answer 22

thank you!! I'm following well so far. Im just expanding them out now to hopefully successfully isolate L3 .... :)

Answer 23

\( L_1L_2 + L_1L_3 =LL_2 + LL_3+LL_1\)

\(L_1L_3 - LL_3 = LL_1 + LL_2- L_1L_2\)

\(L_3(L_1 - L) = LL_1 + LL_2- L_1L_2\)

\(L_3 = \dfrac{LL_1 + LL_2- L_1L_2}{ L_1 - L } \)

Answer 24

Argh you are amazing. I see :D THANK YOU! Far out thank so much !!!!!

Answer 25

Hopefully this approach will help me with the others that Im getting wrong that are similar. It should :) Thanks so much !! If you have time and are up for it can I show you another one I am currently unable to get right? If not I will try and tackle ot with my new tools you have just given me :)

Answer 26

Hi are you still there..? Im sorry but I double checked and the answer is different :( Its;

LL1
L3 = ------ - L2
L1 - L

To clarify, the answer in words is ....

L3 = LL1 divided by L1 - L, minus L2

Answer 27

\(L_3 = \dfrac{LL_1 + LL_2- L_1L_2}{ L_1 - L }\)

\(L_3 = \dfrac{LL_1}{L_1 - L} + \dfrac{LL_2- L_1L_2}{ L_1 - L }\)


\(L_3 = \dfrac{LL_1}{L_1 - L} - \dfrac{L_1L_2 - LL_2 }{ L_1 - L }\)

\(L_3 = \dfrac{LL_1}{L_1 - L} - \dfrac{L_2(L_1- L) }{ L_1 - L }\)

\(L_3 = \dfrac{LL_1}{L_1 - L} - \dfrac{L_2(\cancel{L_1- L}) }{\cancel{ L_1 - L} }\)

\(L_3 = \dfrac{LL_1}{L_1 - L} - L_2 \)

Answer 28

I started from our answer above and changed it into your answer.
As you can see, both are the same just in different forms.

Answer 29

oh boy - give me a sec haha :)

Answer 30

gtg
I'll read your comments tomorrow

Answer 31

Yep, so they are (both the same just in different forms). Thanks very much for showing me all that, its helped me out a lot. THANKS :D

Answer 32

You're very welcome.