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If 15(2x2 – Y2) = 7xy, Find X: Y; If X and Y Both Are Positive. - ICSE Class 10 - Mathematics

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Question

If 15(2x2 – y2) = 7xy, find x: y; if x and y both are positive.

Solution

`15(2x^2 - y^2) = 7xy`

`(2x^2 - y^2)/(xy) = 7/15`

`(2x)/y - y/x = 7/15`

Let `x/y = a`

`∴ 2a - 1/a = 7/15`

`(2a^2 - 1)/a = 715`

`30a^2 - 15 = 7a`

`30a^2 - 7a - 15 = 0`

`30a^2 - 25a + 18a - 15 = 0`

5a(6a - 5) + 3(6a - 5) = 0

(6a - 5)(5a + 3) = 0

`a = 5/6, -3/5`

But a cannot be negative

∴ a = 5/6

`=> x/y = 5/6`

=> x : y = 5 : 6

  Is there an error in this question or solution?

APPEARS IN

 Selina Solution for Selina ICSE Concise Mathematics for Class 10 (2018-2019) (2017 to Current)
Chapter 7: Ratio and Proportion (Including Properties and Uses)
Ex.7D | Q: 8

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Solution If 15(2x2 – Y2) = 7xy, Find X: Y; If X and Y Both Are Positive. Concept: Ratios.
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