Question 472703
Subtract the 2nd equation from the 3rd:  z - y = by - cz, whence z(1+c) = y(1+b).

Similarly, x(1+a) = y(1+b).

==>  x(1+a) = y(1+b) =  z(1+c).
==> {{{y = ((1+a)/(1+b))x }}} and {{{z = ((1+a)/(1+c))x }}}

Subsitute into the 1st equation: {{{x = b*((1+a)/(1+b))x  + c*((1+a)/(1+c))x }}}.
Since x is not identically 0, we can cancel x throughout.

==> {{{1 = b*((1+a)/(1+b))  + c*((1+a)/(1+c))}}}.

==> {{{1/(1+a) = b/(1+b) + c/(1+c)}}}

==> {{{1 - a/(1+a) = b/(1+b) + c/(1+c)}}}

The conclusion follows after one more step.