Question

(y+z)p+(z+x) =x+y

Answer 1

$(y+z)p+(z+x)q =x+y$

$\frac{dx+dy+dz}{2(x+y+z)}=\frac{dx-dy}{y-x}$

$\frac{ln(x+y+z)}{2}=-ln(x-y)+lnc$

$(x+y+z)(x-y)^2=c_1$

$\frac{dx-dy}{y-x}=\frac{dy-dz}{z-y}$

$ln(x-y)=ln(y-z)+lnc_2$

$\frac{x-y}{y-z}=c_2$