- 14 Oct '07 19:09I'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

5((27-3y)/5) + 3y = 27

27-3y +3y = 27

27=27...................why is this happening?

We are studying symmetries in class and and I want to prove that one equation equals or does not equal another,....other than the substitution of numbers into them.

for instance: does 5x-5y=0=-5x+5y,...the check to see if they are symmetric to the origin........in this case they would because x=y making the equation symmetric to the origin

I found this out by saying the addition of the two equations should equal zero 5x-5y+(-5x + 5y) = 0

My question is, by using this method where will I run into complications if any? - 14 Oct '07 19:15You cannot substitute something into the original equation, look at it this way:

x+1=y

so x=y-1

if we substitute into the original you get

y-1+1=y

or y=y

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. - 14 Oct '07 19:245x + 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

x=0, y=9

x=-6,000 y=-10,009

x=5.1, y=0.5

All you can do is describe x in terms of y or vice versa.

And as for whats wrong with proving 27=27 .. nothing! Its correct.

Also another way of looking at simple algebra is to consider the geometry. Plot 5x + 3y = 27 and all solutions are on that line. Add another equation, say 5x + y = 9. Plot that line. Your solution is the intersection of the two lines. (And will agree with your substitution method)

Hope that helps - 14 Oct '07 19:34

so there is a infinate number of solutions....ok*Originally posted by doodinthemood***You cannot substitute something into the original equation, look at it this way:**

x+1=y

so x=y-1

if we substitute into the original you get

y-1+1=y

or y=y

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.

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 - 14 Oct '07 19:40

And you are talking about the composition of functions....correct?*Originally posted by wolfgang59***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

x=0, y=9

x=-6,000 y=-10,009

x=5.1, y=0.5

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

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 - 14 Oct '07 19:48

nevermind 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*Originally posted by joe shmo***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 - 14 Oct '07 19:55

5x - 5y + (-5x + 5y) will always be 0. No matter what 5x - 5y is*Originally posted by joe shmo***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 - 14 Oct '07 20:05

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 TheMaster37***5x - 5y + (-5x + 5y) will always be 0. No matter what 5x - 5y is** - 14 Oct '07 20:08

now 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*Originally posted by joe shmo***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:'(** - 15 Oct '07 02:51

You cannot solve one equation in two variables. You need one equation for each variable to solve the system.*Originally posted by joe shmo***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? - 15 Oct '07 16:21 / 4 editsIf 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) = 1, and 27 mod 1 = 0, so this equation has a solution in whole numbers.

The whole solution is x = x_0 - bk and y = y_0 + ak where x_0 and y_0 are one of the roots for each variable.

In this equation we find than one of the whole solutions is x_0 = 3 and y_0 = 4, so the complete solution is

{x = 3 - 3k; y = 4 + 5k where k is any whole number.

P.s. The solution is meant to be in a system, I just don't know how to write a larger { here.

EDIT: Now, can you solve these?

17x - 13y = 5

6x + 15y = 8

EDIT: Actually there is even an algorithm on finding x_0 and y_0. I think it involves euclidian algorithm on finding G.C.M. and something like that but I don't really remember. No exams on this theme for me anymore. - 15 Oct '07 22:07

I have read your response, but I am undereducated and am having trouble making sense of it.*Originally posted by kbaumen***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.

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 sure I understand that at all along with its y variation. you say that x_0 and y_0 are one of the roots for each variable. What kind of root?

Perhaps If I could get a more solid handle on these operations then I could begin to grasp the concept.

However, I dont expect you to teach me algebra. So if i need to learn to much information just let me know. I dont want anyone to waste their time. I am currently in college algebra and I believe that they will have to tackle these problems sooner or later. If need be I can wait, but thanks to all tose who helped. - 16 Oct '07 09:26 / 2 edits

Here goes;*Originally posted by joe shmo***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.

He might as well have said that x0 and y0 were some numbers. The usage of the word root is alien to me as well here.

mod is short for modulo. 52 modulo 7 is the remainder after dividing by 7; 3 in this case. 16 mod 3 is 1, 6 mod 3 is 0 and anything mod 1 is 0.

GCM is a non-existant term. It's a mixture of GCD, Greatest Common Divisor and SCM, Smallest Common Multiple.

The GCD is the largest number that divides both starting numbers. The GCM of 14 and 18 (denoted as GCD(14, 18)) is 2. GCM(12, 18)=6 and so on. To calculate it for larger numbers one uses Euclid's Algorithm (you can find that on Wikipedia, too long to describe here).

SCM is the smallest number that is a multiple of both starting numbers. So SCM(7,3) = 21, and SCM (12, 16) = 4.

You calculate it by taking the product of the numbers and then divide by their GCD.

Hope that's clear enough