Question

if x and y are 2 positive real numbers such that 9x²+y²=58 and xy=7 then find the value of 3x+y

Answer 1

HELLO DEAR !!!

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GIVEN,

$9 {x}^{2} + {y}^{2} = 58 \: \: \: and \: \: \: xy = 7 \\ \\ to \: find \: = > \: 3x + y$

FIND :-

$9 {x}^{2} + {y}^{2} = 58 \\ \\ = > ( {3x)}^{2} + ( {y)}^{2} + 2 \times 3x \times y - 2 \times 3x \times y = 58 \\ \\ = > ( {3x + y)}^{2} - 6xy = 58 \\ \\ = > ( {3x + y)}^{2} - 6 \times 7 = 58 \: \: \: (using \: xy \: = 7) \\ \\ = > ( {3x + y)}^{2} = 58 + 42 \\ \\ = > ( {3x + y)}^{2} = 100 \\ \\ = > 3x + y = \sqrt{100} \\ \\ = > 3x + y = \: 10$

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Hope it helps u !!!

# Nikky

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Answer 2

Firstly 9x^2 is square of 3x and y^2 is square of y therefore if we add 2(3xy)
And subtract 2(3xy) we will get 9x^2 + y^2. But we'll apply the identity of (a+b) ^2 and we'll get the answer as shown in the attachment. I hope it helped.
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