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Solve: x3+36-x=2(6+x)15;(x≠6) - Mathematics

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Question

Solve:

`x/3 + 3/(6 - x) = (2(6 +x))/15; (x ≠ 6)`

Sum

Solution

`x/3 + 3/(6 - x) = (2(6 + x))/15; (x ≠ 6)`

⇒ `(x(6 - x) + 3 xx 3)/(3(6 - x)) = (12 + 2x)/15`

⇒ `(x(6 - x) + 3 xx 3)/(6 - x) = (12 + 2x)/5`

⇒ `(6x - x^2 + 9)/(6 - x) = (12 + 2x)/5`

⇒ 30x – 5x2 + 45 = 72 + 12x – 12x – 2x2

⇒ 30x – 5x2 + 45 = 72 – 2x2

⇒ 3x2 – 30x + 27

⇒ x2 – 10x + 9 = 0

⇒ x2 – 9x – x + 9 = 0

⇒ x(x – 9) – 1(x – 9) = 0

⇒ (x – 9)(x – 1) = 0

⇒ x – 9 = 0 or x – 1 = 0

⇒ x = 9 or x = 1

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Chapter 5: Quadratic Equations - Exercise 5 (C) [Page 59]

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Selina Mathematics [English] Class 10 ICSE
Chapter 5 Quadratic Equations
Exercise 5 (C) | Q 22.1 | Page 59
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