Is this fractional equation true for all real numbers? Is this equation true for all real values of $a,b,c,x,y,z$?
If $$\frac{a}{x} = \frac{b}{y} = \frac{c}{z} = t$$
Then $$\frac{a+b+c}{x+y+z} = t$$
I've tried experimenting with several values and they make an equality, but does this equation hold true for all real values?
(Almost always) yes: from the first set of equations, you have $a=xt, b=yt, c=zt$. Therefore,
$$\frac{a+b+c}{x+y+z} = \frac{xt+yt+zt}{x+y+z} = t\frac{x+y+z}{x+y+z} = t$$
provided, of course, that $x+y+z\neq 0$.

An example of this: imagine a course where you're graded on homework, a midterm, and an exam. If you get the same grade on each of these components, then that's your final grade, regardless of how they're weighted.
$ \text{Not true for all real numbers}  $
Comparing both equations,
$ \frac{a}{x}+\frac{b}{y}+\frac{c}{z}= \frac{a+b+c}{x+y+z} $
$ (x+y+z)\times (\frac{a}{x}+\frac{b}{y}+\frac{c}{z})=a+b+c $
$ \frac{1}{x} (ax+ay+az) + \frac{1}{y}(bx+by+bz) + \frac{1}{z}(cx+cy+cz) = a+b+c$
$ a+\frac{ay+az}{x} +b+ \frac{bx+bz}{y} + c+ \frac{cx+cy}{z}=a+b+c $
$a+b+c= a+b+c +(\frac{ay+az}{x} + \frac{bx+bz}{y} + \frac{cx+cy}{z}) $
$ \frac{ay+az}{x} + \frac{bx+bz}{y} + \frac{cx+cy}{z} = 0 $
$ \text{As} $ $ x,y,z \neq 0  $
one Solution is when  $ a=b=c=0 $, and have other solutions as well but Not true for all Real Numbers.
Thanks to