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?
On
$ \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
On
Alternative approach that is somewhat inferior to the approach taken in Theophile's answer.
It is assumed that none of $a,b,c,x,y,z, (x + y + z)$ are equal to zero.
Without loss of generality, there exist (Real non-zero) scalars $r,s$ such that $a = rb$ and c = sb$.$
Therefore, $\frac{rb}{x} = \frac{b}{y} \implies \frac{r}{x} = \frac{1}{y} \implies x = ry$. Similarly, $z = sy.$
Therefore, $\frac{a + b + c}{x + y + z} = \frac{b(r + 1 + s)}{y(r + 1 + s)}.$
Note that the assumption that $(r + 1 + s) = 0$ would imply that
$(x + y + z) = y(r + 1 + s) = 0$, which is a contradiction.
Therefore, you know that
$\frac{b(r + 1 + s)}{y(r + 1 + s)} = \frac{b}{y} = t.$
(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.