how to prove $ax + by = cx + dy \implies a = c, b = d$?

Question

Actually the question is in the title. I just have saw such a method

$$ ax + by = cx + dy \implies a = c, b = d $$

in my textbook, so I can assume it is true, but I'm very interested on proving this, and failed to do this myself. Could anyone explain this, please.

Thank you.

Answer 1

If this equality holds for all $x$ and $y$ then

• $x=1$ and $y=0$ gives $a=c$
• $x=0$ and $y=1$ gives $b=d$.

Answer 2

$ax + by = cx + dy \implies (a-c)x+(b-d)y=0$

if we assume that neither $x$ nor $y$ is zero and they don't have any know relation with each other (for example if $x=-y$ then $a-c=b-d$) then we can conclude that $a=c$ and $b=d$

Answer 3

The equality does not need to hold for all real x and y, i.e. the domain of the values of x and y might be a subset of the real numbers, or imaginary numbers or whole numbers etc. The important thing is that the values should be independent. If x and y can take on values independently then we can have two cases:

First choose any arbitrary value of x and y and find the relations between a,b,c and d. Then keeping the value of x the same vary the value of y for the second equations relating a,b,c and d. This will give the result b=d.

Similarly do the same keeping the value of y the same and varying the value of x and it will lead to the result a=c.