math help: Can anyone figure this out? - Help.com

GOOD LUCK! I cant even start to understand

a+b+c+d=771
2a=3b
2a/3=b
5(2a/3)/4=c —> (10a/15)/4=c
5(10a/15)/6=d –> (50a/75)/6=d

Working on it … I vaguely remember this … there is someway you use all three equations. But, **** it is like 1:30 AM … give me a second kid.

its simple actually..
2a=3b
5b=4c
5c=6d
since 2a=3b, b =2a/3
similarly c= 5b/4which also means c= 5/4 (2a/3)= 5a/6
d = 5c/6 = 25/36a
now substitute all these and u will get a equation containing only a

in my previous post read the last line as d= 25a/36 . sorry for the typo

I get b=181.411764, a=114.113851992
working on the last two…
Dracoreds tell me if that is what you get. It has been forever since I have done one of these.

well the values come to a= 241.35
b= 160.9
c= 201.13
d= 167.6

Yeah … I thought I screwed up somewhere. This is a refresher

Can you write out how you got that? I keep screwing this up and am getting frustrated with myself. I am sure you are right, because your numbers fit.

u trying to get ur homework done without effort on
ur part??
:)

well..the equation after substitution comes out to
a + 2a/3 + 5a/6 + 25a/36 = 771

or 115a/36 = 771
or a = 241.35
am sure u can work out the rest.

Not my homework … It was the person who posted this, but I was going to try and help him with the problem. Now I am frustrated with myself.

I end up with 110a/36=771. Am I adding wrong?

a+(2a/3)+(5a/6)+(25a/36) –> (36a/36)+(24a/36)+(25a/36)+(25a/36) –>
(36a+24a+50a/36) =771

yeah..it shud be (36a/36)+(24a/36)+(30a/36)+(25a/36)

OMG! I AM SUCH AN IDIOT! I will blame it on the time here … but ****. I can pass Calc., but I screw something like that up …

SCREAMING IN MY MIND>

5*6=30 … DUH!

it happens… but u were on the right track… was fun solving it

Yeah, it has been awhile since I have seen one of those.

Isn’t there another way where you can add the different equations together or something? (Do you know what I am thinking of?)

there can be many ways.. but they will essentially boil down to reducing all the variables to a single one in the end.

Oh wait, I remember what I am thinking of…

4x+3y+2z= Whatever
2x+3y+5z= Whatever

And then you can find out what x equals between both equations by adding both equations together and then solving for each variable one way or another.

Ringing a bell at all … I don’t think these are the voices in my head talking again … I just vaguely remember this.

even in that.. u will have to first substract the two resulting in an equation with only 2 variables.. then substitute again to get it to one variable..same thing
(4x+3y+2z)- (2x+3y+5z) = whatever
resulting in 2x-3z = whatever

Yes exactly … well thank you for the help. It is really true, if you don’t use it, you lose it. I should be better about keeping my mind in better shape.

play chess…. :D
just kidding

rizwanazim…. u have got it all wrong.

a/b comes to 3/2 and not 2/3.
moreover going by ur calculations.. a= 150, b=225, c= 180, and d= 270
the total doesnt come to 771.