If a = \(\frac{xy}{x+y}\), b = \(\frac{xz}{x+z}\), and c = \(\frac{yz}

1.	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» (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 = \(\frac{2abc}{(ac+bc-ab)}\)

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»

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