Given: (x+y+z)=k (x+z−y)×(x−z+y)∝yz Formula used: Algebraic manipulation and properties of proportionality. From (x+y+z)=k, we can derive: x+z−y=(x+y+z)−2y=k−2y x−z+y=(x+y+z)−2z=k−2z Calculations: Given that (x+z−y)×(x−z+y)∝yz ⇒(k−2y)×(k−2z)=C×yz (where C is the constant of proportionality) ⇒k2−2kz−2ky+4yz=C×yz ⇒k2−2k(z+y)+4yz=C×yz ⇒k2−2k(y+z)=C×yz−4yz ⇒k2−2k(y+z)=(C−4)yz Now, let's rearrange to find what (y+z−x) is proportional to. From (x+y+z)=k, we have y+z=k−x. Substitute (y+z) in the equation k2−2k(y+z)=(C−4)yz : ⇒k2−2k(k−x)=(C−4)yz ⇒k2−2k2+2kx=(C−4)yz ⇒−k2+2kx=(C−4)yz ⇒k(2x−k)=(C−4)yz We are interested in (y+z−x). We know (y+z−x)=(k−x)−x=k−2x From k(2x−k)=(C−4)yz, we can write: ⇒−k(k−2x)=(C−4)yz ⇒k−2x=
(C−4)yz
−k
⇒k−2x=
(4−C)yz
k
Since k,C are constants,
(4−C)
k
is also a constant. Therefore, ( k−2x ) is proportional to yz . Since (y+z−x)=k−2x, ∴(y+z−x) is proportional to yz.
