Question

If x + y ∝ z when y is constant and z + x ∝ y when z is constant, show that x + y + z ∝ yz when both y and z vary.

Answer 1

Answer:

Given,

x∝y

Then x=ay.....(1) [ Where a is proportionality constant]

Also, y∝z

or, y=bz......(2) [ Where b is proportionality constant]

Using (2) in (1) we get,

x=ay=abz

or, x∝z. [ Since ab is proportionality constant]

Answer 2

★ GivEn :-

• x + y ∝ z when y is constant
• z + x ∝ y when z is constant

★ To ProVe :-

• x + y + z ∝ yz when both y and z vary.

★ SoluTiOn :-

➟ 1st Step :-

➬ x + y ∝ z ( y is constant )

or, x + y = kz ( k is assumed to be constant )

or, x + y + z = kz + z ( adding z in both sides )

or, x + y + z = z ( k + 1 ) _____________ i)

➟ 2nd Step :-

➬ z + x ∝ y (z is constant)

or, z + x = k'y ( k' is assumed to be constant )

or, x + y + z = k'y + y ( adding y in both sides )

or, x + y + z = y ( k' + 1 )______________ii)

➟ Final Step :-

➬ Applying the joint variation theorem to i) and ii) we get,

x + y + z ∝ yz [as ( k + 1 ) and ( k' + 1 ) is constant]

➬ Hence, it is Proved.

➟ Joint Variation Theorem :

According to joint variation theorem, if x ∝y, when z is constant and x ∝ z, when y is constant then x ∝yz when both y and z vary.