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This is because is,

$$\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$$

then,

$$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{a+c+e}{b+d+f}$$

This can be proved as follows.

Let $$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = r$$

$$\implies a = rb, c = rd, e = rf$$

$$\frac{a+c+e}{b+d+f} = \frac{r(b+d+f)}{b+d+f} = r$$

So, $$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{a+c+e}{b+d+f}$$

Here, $$r = \frac{y+x+y+y}{x-z+z+y} = \frac{2(x+y)}{x+y} = 2, (x+y\not=0)$$

Compare $r$ with every fraction and obtain $x:y:z$

19aksh
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  • Is this the same for a/b = c/d = e/f = a-c-e/b-d-f? – Kaushik Palavalasa Jun 09 '19 at 15:35
  • Yes you're correct *but be sure it's not zero* as $\frac{0}{0}$ is undefined. For e.g, $\frac{3}{4} = \frac{2}{4} = \frac{1}{2} =0.5$ but $ \frac{3-2-1}{6-4-2} = \frac{0}{0}$ is undefined. – 19aksh Jun 09 '19 at 15:55
  • Is it similar for $$a/b + c/d + e/f = g/h + i/j + k/l$$ where it is equal to $$a+c+e+g+i+k/b+d+f+h+j+l$$ – Kaushik Palavalasa Jun 10 '19 at 17:59
  • Yes here too the sum shouldn't be equal to zero, they've mentioned this in the solution by saying it is true unless $x+y = 0$ – 19aksh Jun 10 '19 at 18:01