If $\,ax+b=cx+d,\,$ then is $\,a=c\,$ and $\,b=d$? I am a high school student my maths teacher said that if $\,ax+b=cx+d,\,$ then is $\,a=c\,$ and $\,b=d.\,$ Can someone give me a prove of this?
 A: Let $x=0$, then $b=d$. So $ax+b=cx+b$. So $ax=cx$. Then let $x=1$ to get $a=c$
A: The statement "ax + b = cx + d implies a=c and b = d" is not true and should be obviously so.  Simply do $b = cx + d - ax$ and you get $ax + b = ax + cx + d - ax = cx + d$.  $a, b, c$ and $x$ can be anything you like.
That's silly.
HOWEVER the statement "ax + b = cx + d for all possible values of x (where x is not a constant) implies a=c and b = d" is true. 
Pf:
$ax + b = cx + d \iff$
$(a-c)x = d - b$.
$d- b$ is a constant value.  If $(a-c)\ne 0$ then $(a-c)x$ can have multiple values for different values of $x$.  As $(a-c)x$ is constant for all possible values of $x$, the only way this is possible is if $(a-c)  = 0$ i.e. a = c.  
But then $(a-c)x = d-b \implies 0x = 0 = d-b \implies b = d$.
(I can't say: Let $b = cx + d - ax$ as $cx +d -ax$ will have different values for different values of $x$.  ... unless $c =a$ ... in which case I'm saying "Let $b = d$"....)
A: By the fundamental theorem of algebra (FTA), a first degree polynomial has exactly $1$ root.  Thus if there are $2$ or more values of $x$ for which the equation holds, then $(a-c)x+(b-d)$ is the zero polynomial.   So $a=c$ and $b=d$.
