If: $$3x+2y=mx+ny$$
can $m$ and $n$ take other values than $3$ and $2$ respectively, for any real values for $x$ and $y$ except $(x,y)=(0,0)$?
If no, can you please give an easy to understand proof for a $12$th grade student? Is this something related to Equating coefficients?
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(Of course we need to assume that $m$ and $n$ are required to be integers or the problem is utterly trivial.)
Hint: Say $x=1$ and $y=\sqrt 2$. If $3x+2y=mx+ny$ that says $(2-n)\sqrt 2=m-3$. If $n\ne 2$ then $\sqrt 2=(m-3)/(2-n)$, which is impossible because...
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Note that
for $(x,y)=(1,0)\implies 3\cdot 1+2\cdot 0=m\cdot 1+n\cdot 0 \implies m=3$
for $(x,y)=(0,1)\implies 3\cdot 0+2\cdot 1=m\cdot 0+n \cdot 1 \implies n=2$
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No, they can't. Yes, it is the same as equating coefficients. Any non-zero polynomial over a field has only finite roots, and so does $(3-m)x+(2-n)y$. If this equals zero for every real number, that means it has infinite solutions, and thus is the zero polynomial (the one whose coefficients are all zero). The proof of this theorem is usually done by induction on the polynomial's degree.
It depends on which values $m$ and $n$ can have. If you are searching for numbers, $m=3$ and $n=2$ are the only solutions.
More generally, you can write your equation to $mx = 3x + 2y - ny$, which is equivalent to $m=3 + \frac{2y-ny}{x} = 3 + \frac{(2-n)y}{x}$ by dividing by $x$. So you have to exclude $x=0$. Now you can choose an arbitrary $n$ and for a solution you have got to "fix" your $m$ with the formula above. Back to the beginning. If you are searching numbers, then you don't want that your solution for $m$ and $n$ depends on $x$ and $y$, so the second summand $\frac{(2-n)y}{x}$ has to be $0$, which happens, when $2-n=0$, so if $n=2$ and tehrefore $m=3$.