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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?

William
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3 Answers3

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(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.

Théophile
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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

$\frac{b(r + 1 + s)}{y(r + 1 + s)} = \frac{b}{y} = t.$

user2661923
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$ \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

Atique Ahmed
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  • The problem was updated; the question is different now. But even for the original question, your argument doesn't make sense. How can you conclude that $a = a + \frac{ay+az}x$, for example? There are many solutions apart from $a=b=c=0$. – Théophile Apr 23 '21 at 00:54
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    @Théophile How this problem has many solutions? I just conclude $ a= a+ \frac{ay+az}{x} $ to make him understand.. other wise we will conclude from equation, $ a+b+c (\frac{ay+az}{x} + \frac{bx+z}{y} + \frac{cx+cy}{z}) = a+b+c $ . we will get $ \frac{ay+az}{x} + \frac{bx+z}{y} + \frac{cx+cy}{z}=0 $ we can't make $ x,y,z \in {0} $. – Atique Ahmed Apr 23 '21 at 12:54
  • Here's a nice integer solution: $$\frac1{-1}+\frac21+\frac33 = \frac{1+2+3}{-1+1+3}.$$ To see that there are many more solutions, fix $a,b,c,z,t$, for example. You now have two equations in $x$ and $y$: one is effectively a hyperbola, and the other, a line. I encourage you to graph this to see what happens. – Théophile Apr 23 '21 at 16:34
  • Or, if you like, take your last equation and multiply by $xyz$. This gives $ayz(y+z)+bxz(x+z)+cxy(x+y)=0$. Again, fix $a,b,c,z$. You now have a curve in $x,y$. (And again, I encourage you to graph this to see for yourself.) – Théophile Apr 23 '21 at 16:40