Originally posted by doodinthemoodso there is a infinate number of solutions....ok
You cannot substitute something into the original equation, look at it this way:
if we substitute into the original you get
If you mess about a bit you could also get that to say 1=1. This is because in any equation with two variables, the numbers can be anything. X+1=y cannot be solved because it just means y is 1 bigger than x.
Originally posted by wolfgang59And you are talking about the composition of functions....correct?
5x + 3y = 27
Golden rule; if you have N variables you need (at least) N equations to solve uniquely (or other clues like x & y are integers).
Without further info your equation has infinite solutions, eg
All you can do is describe x in terms of y or vice versa.
And as for whats wrong with proving ...[text shortened]... section of the two lines. (And will agree with your substitution method)
Hope that helps
Originally posted by joe shmonevermind about the compositions of functions bit.......In retrospect I dont believe thats what you were talking about.....you were talking about intersecting lines....you saying that the point of intersection is the solution for both line equations.......got ya...i suppose
And you are talking about the composition of functions....correct?
It has helped me.......Its good to have a little reassurance that I am incorrect, because I should be studying for Chemistry.....haha
well thank you also
Originally posted by joe shmo5x - 5y + (-5x + 5y) will always be 0. No matter what 5x - 5y is
so there is a infinate number of solutions....ok
and I dont think that I picked a good number to experiment with
I see that while 5x-5y+(-5x+5y)= 0
using that same logic would get me nowhere if 5x-5y=24
giving me 5x-5y+(-5x + 5y)= 48
Thanks for the imput
Originally posted by TheMaster37yes, my logic for a proof was that 0=0. when making the equation equal to 24,then adding itself to it self should =48, and I expected to see 48=48, as in 0=0...........but i didnt....haha:'(
5x - 5y + (-5x + 5y) will always be 0. No matter what 5x - 5y is
Originally posted by joe shmonow rereading you post I understand, that you understood what I meant......hence I did not have to reexplain it to you...........I'm a doofass
yes, my logic for a proof was that 0=0. when making the equation equal to 24,then adding itself to it self should =48, and I expected to see 48=48, as in 0=0...........but i didnt....haha:'(
Originally posted by joe shmoYou cannot solve one equation in two variables. You need one equation for each variable to solve the system.
I'm new to the wonderful world of algebra and I have a question
equation 5x + 3y = 27 I have never solved equations like this and have not been taught how, but I can prove that 27=27,but i cant solve for the variables;at least not using algebra
here is what I do
solve for x in terms of y so, x=(27-3y)/5 and plug that into the original equation
...[text shortened]... x + 5y) = 0
My question is, by using this method where will I run into complications if any?
Originally posted by kbaumenI have read your response, but I am undereducated and am having trouble making sense of it.
If your equation has a criteria, that it must be solved in whole numbers, than the solution isn't too difficult to find.
5x + 3y = 27.
Let's make it general
ax + by = c
This equation has a solution only in whole numbers only if c divides with G.C.M.(a; b).
(A theorem I had to remember while learning number theory last year)
Here G.C.M.(5;3) ...[text shortened]... and something like that but I don't really remember. No exams on this theme for me anymore.
Originally posted by joe shmoHere goes;
I have read your response, but I am undereducated and am having trouble making sense of it.
here is what information I need to continue my persuit
What is a G.C.M? I think it means Greatest Common Multiple; am I wrong? More importantly, how is it derived and why is it significant?
next, 27 mod 1 = 0. What does mod mean?
x=x_0-bk Im not s ...[text shortened]... oblems sooner or later. If need be I can wait, but thanks to all tose who helped.