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Question
If (3x² + 2y²) : (3x² – 2y²) = 11 : 9, find the value of `(3x^4 + 5y^4)/(3x^4 - 5y^4)`
Solution
`(3x^4 + 5y^4)/(3x^4 - 5y^4) = (11)/(9)`
Applying componendo and dividendo
`(3x^2 + 2y^2 + 3x^2 - 2y^2)/(3x^2 + 2y^2 - 3x^2 + 2y^2) = (11 + 9)/(11 - 9)`
⇒ `(6x^2)/(4y^2) = (20)/(2)`
⇒ `(3x^2)/(2y^2)` =
⇒ `x^2/y^2 = 10 xx 2/3`
= `(20)/(3)`
`(3x^4 + 5y^4)/(3x^4 - 5y^4)`
= `((3x^4)/(y^4) + (25y^4)/(y^4))/((3x^4)/(y^4) - (25y^4)/(y^4)`
= `(3(x^2/y^2)^2 + 25)/(3(x^2/y^2)^2 - 25)`
= `(3 xx (2/3)^2 + 5)/(3(20/3)^2 - 25)`
= `(3 xx 400/9 + 25)/(3 xx 400/9 - 25)`
= `(400/3 + 25/1)/(400/3 - 25/1)`
= `((400 + 75)/(3))/((400 - 75)/(3)`
= `(475)/(3) xx (3)/(325)`
= `(19)/(13)`.
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