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Using Componendo and Dividendo, Find X : Y - Mathematics

Sum

Given `("x"^3 + 12"x")/(6"x"^2 + 8) = ("y"^3 + 27"y")/(9"y"^2 + 27)`

using componendo and dividendo, find x : y

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Solution

`("x"^3 + 12"x")/(6"x"^2 + 8) = ("y"^3 + 27"y")/(9"y"^2 + 27)`

Applying componendo and dividendo, we get 

`("x"^3 + 12"x" + 6"x"^2 + 8)/("x"^3 + 12"x" - 6"x"^2 - 8) = ("y"^3 + 27 + 9"y"^2 + 27)/("y"^3 + 27"y" - 9"y"^2 - 27)`

⇒ `("x"^3 + 3(1)(4)"x" + 3(1)(2)"x"^2 + 2^3)/("x"^3 + 3(1)(4)"x" + 3(1)(2)"x"^2 + 2^3) = ("y"^3 + 3(1)(9)"y" + 3(1)(3)"y"^2 + 3^3)/("y"^3 + 3(1)(9)"y" + 3(1)(3)"y"^2 + 3^3)`

⇒  `("x"^3 + 3(1)(4)"x" + 3(1)(2)"x"^2 + 2^3)/("x"^3 + 3(1)(2)"x" + 3(1)(4)"x"^2 + 2^3) = ("y"^3 + 3(1)(9)"y" + 3(1)(3)"y"^2 + 3^3)/("y"^3 + 3(1)(3)"y" + 3(1)(9)"y"^2 + 3^3)`

⇒ `("x" + 2)^3/("x" - 2)^3 = ("y" + 3)^3/("y" - 3)^3`

Again applying componendo and dividendo, we get

`("x" + 2 + "x" - 2)/("x" + 2 - "x" + 2) = ("y" + 3 + y - 3)/("y" + 3 - "y" + 3)`

⇒ `(2"x")/4 = (2"y")/6`

⇒ `"x"/2 = "y"/3`

Applying alternendo, we get

`"x"/"y" = 2/3`

⇒ x : y = 2 : 3

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APPEARS IN

Selina Concise Maths Class 10 ICSE
Chapter 7 Ratio and Proportion (Including Properties and Uses)
Exercise 7 (D) | Q 23 | Page 103
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