This is from a competition math problem I had recently that I just couldn't figure out.
If $ (x+y):(y+z):(x+z) = 1:2:4$ and $x+y+z=35$ compute the value of x.
I can tell that $7*(x+y)=2x+2y+2z$ which can be simplified as $5x+5y=2z$
Also $7/2(y+z)=2x+2y+2z$ which can be written as $1.5z+1.5y=2x$
And finally $7/4(x+z)=2x+2y+2z$ or $-0.25x-0.25z=2y$
Then by simple replacement algebra you can find that $4.375y-0.625z=2z$ or $4.375y=2.625z$ or $y=0.6z$ this makes the 2nd formula into $2.4y=2x$ or $1.2y=x$
With all of these in mind, $x+y+z=(6/5)y+y+(5/3)y= (58/15)y=35$ so $y=525/58$ since $1.2y=x$ we can tell that x is $2625/348$
This answer may or may not be right (I never got the answers) but in itself the method is too complicated. I was supposed to do this in less than 10 minutes. Is there a faster or simpler way to do this that I just missed?
If you are attuned to the cyclic symmetry of things, you might notice it can be rewritten
$$(x+y):(y+z):(z+x)=1:2:4\qquad\text{and}\qquad (x+y)+(y+z)+(z+x)=70$$
which gives $(x+y)+2(x+y)+4(x+y)=70$, or $x+y=10$. From this it follows that $y+z=2(x+y)=20$, and we can finish by invoking $x+y+z=35$, which gives $x=15$.