Take.
3x=1y
Loosen out the x.
The standard way is either to devide both sides of the equal sign with 3
The other way which gives exactly the same answer is to move the 3 to the other side and changing the sign to a devision.
My way to operate is to put -2x on the both sides of the equalsign. Giving the operation 3x=1y
3x-2x=1y-2x
1x=1y-2x
lets define x and y were x is defined as 2 and y is defined as 5. giving the equation the following look.
3*2=1*5
2=5-4
This is the solution using my technique to solve an equation.
But there is a missing link as you can see...
3*2-2*2 = 1*5-2*2
6-4=5-4
What i have done here is redefined the number 5 to a 6 and 2 to a 1 using an unorthodox technique of solveing an equation. But we don't do that because that is just plain silly, we need to go further
To get it stright, i need to move 6 to the other side of the equal sign and changing the operator to a - sign and finally add a -1 to the side without numebers giving the answer 5-6=-1
I chose to operate diffrently and use other techniques other then the standards. Here i had to add a -1 for the equation to turn out right.
Or in short you altered the equation in a way that isn't mathematically correct. You didn't add -1, since adding minus one would have given you:
5-6-1=-1 which would be -2=-1 and therefore is wrong, in reality you altered the values of the equation on only one side, making it a different equation altogether.
What you have proven with that is, that x=2 and y=5 are valid for 3x-1=1y and that they are not valid for 3x=1y.
But atleast i knew what I was doing. 5-6=-1 is a solution to 3x=1y when the variables are defined in such manners i wished to define it and i add -1. Lets call it for a missing link were you need to imagine a number so hard the equation turns out to be alright!!
Yes, if you alter the equation you can prove everything you want, I can prove x=3, y=5 is correct or x=4, y=5 etc. etc. in short I can prove that any number from positive to negative infinity for x or y is a possible solution as long as I add a certain amount to the equation in the way you have done it.
In fact here is the equation that can be used to calculate the alteration you need to make:
y=3x+t (t being the alteration)
t=y-3x if we now insert x= 2 and y=5 we get
t=5-6=-1
and for some other values
x,y
3,5:t=-4
4,5:t=-7
-259,135:t=912
That is not an unorthodox method but a plain and simple incorrect one for this equation, except if the question was: How do I need to alter the equation 3x=1y in order for (2,5) to be valid?
3x=1y if we define x as 2 and y as 5, which is fully possible because it is a variables, we will however get an error on the calcualtor, but that is because 5 is not a 6 and 6 is not a 5 unless we become silly.
Yes you can define the variables however you like, that doesn't mean that those variables will make any sense in the given context(equation).
But lets push it further. And define x as 2a:s and y as 5b:s. Since we have defined x and y to other variables the equation turns out to be correct when solving it. Even though using an unorthodox or foreign way to operate the equation...
3*2a=1*5b
6a=5b
6a-2x=5b-2x
6a-4a=5b-4a
Still correct.
Fully possible, without any objections. As we have defined the variables x and y to other variables the solution of the equation turns out alright. Using this unorthodox technique to solve the equation will give those numbers. And i am aware that the answer is diffrently from the orthodox way to solve it. But that doesn't change the fact that my way to operate will give that outcome...
Totally correct way of doing it, no objection except for one: What is the solution? You have just artificially added 2 new equations, namely:
x=2a and
y=5b
and put them into the given equation, but that isn't the solution, it can be a way to find the solution but it isn't one itself. The last one tried to find a solution, but this one just stops somewhere in the middle.
This equation 3x=1y defines a line, try to draw it and then draw the point (2,5) and see if it is on the line, if not then x=2 and y=5 is not a possible solution for 3x=1y.
I hope I didn't misread or misinterpreted anything in your post but that is how I see your unorthodox solution.