If a = (xy)/(x+y), b = (xz)/(x+z), and c = (yz)/(y+z), where a, b and c are non-zero, then what is x equal to ?

Question

If a = \(\frac{xy}{x+y}\), b = \(\frac{xz}{x+z}\), and c = \(\frac{yz}{y+z}\), where a, b and c are non-zero, then what is x equal to ?

(a) \(\frac{2abc}{ac+bc-ab}\)

(b) \(\frac{2abc}{ab-ac+bc}\)

(c) \(\frac{2abc}{ab+bc+ac}\)

(d) \(\frac{2abc}{ab+ac-bc}\)

Answer 1

(a) \(\frac{2abc}{ac+bc-ab}\)

Given, c = \(\frac{yz}{y+z}\) ⇒ cy + cz = yz ⇒ yz - cz = cy ⇒ z (y - c) = cy 

⇒ z = \(\frac{cy}{y-c}\)

Also b = \(\frac{xz}{x+z}\) ⇒ z = \(\frac{bx}{x-b}\)

∴ \(\frac{cy}{y-c}\) = \(\frac{bx}{x-b}\) ⇒ cyx - cyb = bxy - bxc

⇒ cyx – cyb – bxy = – bxc 

⇒ – y(bx + bc – cx) = – bxc

⇒ y = \(\frac{bxc}{bx+bc-cx}\)

Now, a = \(\frac{yz}{y+z}\) ⇒ y = \(\frac{ax}{x-a}\)

∴ \(\frac{bxc}{bx+bc-cx}\) = \(\frac{ax}{x-a}\)

⇒ abx² + abcx – acx² = bx²c – abcx 

⇒ 2abcx = x² (bc + ac – ab) ⇒ x = \(\frac{2abc}{(ac+bc-ab)}\)