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Consider the system:

$$ ax+by=c \qquad (1) $$ $$ a'x+b'y=c' \qquad (2) $$

For $x$ to be a solution of equation $(1)$, it has to respect the relation $x=(c-by)/a$. We substitute this value of $x$ into the second equation for it to to be solution of both equations; we find $y$. Is this enough proof or do I have to think deeper?

Also another question should I think in maths, because we have a we have b, or do a to have b like in substituion we get in a an equation a relation for x( to be solution of that equation or because it is solution to that equation) and should I think we replace the second equation by this relation ( to have the both equation have the same couple solution or because they have the same couple soulution (y=y,x=x) I am allowed to replace)

Brian Tung
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Selim Jean Ellieh
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  • Please consider using MathJax (and punctuation...) because this is very hard to read. – Klaus Apr 01 '19 at 16:33
  • https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Klaus Apr 01 '19 at 16:36
  • The equation $$x=\frac{c-by}{a}$$ holds only if $$a\neq 0$$ – Dr. Sonnhard Graubner Apr 01 '19 at 16:37
  • OP: I have edited the first half of your question. I think I have captured your intent, but you should check to make sure. (The most substantive change I made was to "fix" the second $x$ in Equation $(2)$ to a $y$.) I don't understand your second part, so I haven't touched it. – Brian Tung Apr 01 '19 at 16:47
  • dont mind the second part thank you, can you just answer is that enough proof why it works as a grade 9 student? – Selim Jean Ellieh Apr 01 '19 at 16:49
  • The first part of your question is a valid logical argument when $a \ne 0$. That case requires special treatment. The last paragraph makes no sense to me. – Ethan Bolker Apr 01 '19 at 17:22

1 Answers1

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If you toss Henry into a meat grinder and get PateofHenry. And Henry is the world's only French Speaking duck. Then tossing a French Speaking duck into a meat grinder is the same thing as tossing Henry into a meat grinder. So the result you get from tossing a French Speaking duck into a meat grinder is the same thing as tossing Henry into a meat grinder because Henry and the French speaking Duck are the exact same thing.

That's all the deepity to it that there is.

$ax + by = c$ means they are are the same thing.

So $ax + by - by = c - by$. Becauxe $ax+by$ and $c$ are the same thing so whatever we do to either of them will be the same thing as doing it to the other one of them. Instead of tossing them into a meat grinder we are subtracting $by$ from them. The result will be the same because they are the same thing.

$by - by = 0$ because we know that every number has an additive inverse that when added to itself is $0$. So $by - by$ is the same thing as zero whatever we say about $by-by$ we can say about $0$ because they are the same thing.

$ax + by - by = c - by \implies ax + 0 = c-by$. And as $anything + 0 = anything$ (that's what $0$ is) we know $ax = c-by$.

Which means $ax = c-by$ are the same thing. So we whatever we do to one is the same thing as doing to the other.

$\frac {ax}a = \frac {c-by}a$. And as $\frac {\cdot}a$ means "undoing" multiplying by $a$ (we know that we can do that because we know numbers work that way; for any non-zero $a$ we can divide by $a$ and when we divide $a$ by $a$ we get $1$ and because division distributes we have $\frac {ax}a = \frac aa x$ so $\frac aa x = 1*x = x$.)

So we know $x = \frac {c-by}a$ so, again $x$ and $\frac {c-by}a$ are both the same thing. So whatever we do to each of them will result in the same result.

So $a'(x) + b'y$ and $a'( \frac {c-by}a) + b'y$ will both be the same thing. (Because $x$ and $\frac {c-by}a$ are the same thing.)

And we know that $a'x + b'y = c'$ means $a'x + b'y$ and $c'$ are the same thing. ANd $a'x + b'y$ and $a'( \frac {c-by}a) + b'y$ are the same thing. So we have three names for the same thing: $c'$ and $a'( \frac {c-by}a) + b'y$ and $a'x+b'y$.

SO $a'( \frac {c-by}a) + b'y$ and $c'$ are two names for the same thing so:

$a'( \frac {c-by}a) + b'y = c'$. And ... we keep going.

No teacher on the planet is going to want this much detail. And no student who doesn't ask when want us explaining in this much detail.

But all good students will at some point ask "How do we know something is true". And in this case the answer is. $M = W$ means $M$ and $W$ are two names for the same thing. ANd what ever we do to $M$ will be the same thing as doing it to $W$ because $M$ and $W$ are the same thing. And that's really all there is to it.

fleablood
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  • Thanks you bro but I think about it in alittle different way i think more about it graphically how x has to verify a relation with y in equation (1) then we force "it" to verify it in equation then we get same x and y up and down is that another good perspective to look at it? – Selim Jean Ellieh Apr 01 '19 at 17:44
  • That's a great perspective to like at it! $ax+by=c$ has an infinite number of solutions (for $x$ and $y$, we are assuming $a,b$ are constants). If we graph these solutions they form a line (we have to build a time machine and ask Rene Descartes how he came up with the idea but assuming concepts of geometry and consistency of arithmetic this makes sense once we have the relationship pointed out to us). $a'x+b'y=c'$ is another line. Assuming they aren't parallel they will have one point of intersection. – fleablood Apr 01 '19 at 18:29
  • But I took your question to be "why can we do substitution?" And the answer to that is equal things are the same. And by substituting the same $ax + by$ and $c$ are the same then $\frac{(ax+by)-by}{a}$ and $\frac {c-by}a$ are the same. And $\frac{(ax+by)-by}a$ and $x$ are the same. So $x$ and $\frac{c-by}a$ is the same. Which means $x$ has a relation with $y$. – fleablood Apr 01 '19 at 18:33
  • thanks you so in maths you can look at why something work in many percpectives :) – Selim Jean Ellieh Apr 01 '19 at 19:12