Solve for $y$

Let :-
$\implies \dfrac{ y}{a-c} = \dfrac x{b-c} = \dfrac z{a-b}=k$

$y=ka-kc \ \ \ ...(1)$

$x=kb-kc \ \ \ ...(2)$

$z=ka-kb \ \ \ ...(3)$

$\implies (1)-(2)-(3)$

$y=\boxed{x+z}$

x=y $\frac{b-c}{a-c}$ = $\frac{yb-yc}{a-c}$ z=y $\frac{a-b}{a-c}$ = $\frac{ya-yb}{a-c}$ x+z= $\frac{y(a-c)}{a-c}$ x+z=y

let a – c =p, b – c =q, a – b = r, then p = q + r, y/p = x/q = z/r = k, then y=kp, x=kq, z=kr. So y=x+z