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

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?

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.

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 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
🙂

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.

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
🙂

I dont think there is any benefit in considering them as functions but if you wanted you could rewrite your equation 5x + 3y = 27 as

y = 27-5x/3

and consider the function

f(x) = 27-5x/3

as I said no benefit just a slightly different way of looking at it.

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🙄

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

Originally posted by TheMaster37
5x - 5y + (-5x + 5y) will always be 0. No matter what 5x - 5y is 🙂

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:'(

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?

Originally posted by joe shmo
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😉

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) = 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.

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.

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.
🙂🙂😉😉😉😉