Question

if x and y be two positive real number such that 9x^2+ y^2=96 and xy = 8 then find the value of 3x+y​

Answer 1

Answer:

$9x {}^{2} + y {}^{2} = 96 \: and \: xy = 8 \: given \\ so \: ( 3x + y) {}^{2} = (9x {}^{2} + y {}^{2}) + 6xy \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 96 \: + \: 6 \: \times \: 8 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \:144 \\ thus \: (3x \: + \: y) {}^{2} = \: 144 \\ hence \: \: 3x \: + \: y \: = \: \sqrt{144 } \\ ie \: \: \: \: \: \: \: \: \: 3x \: + \: y \: = \: 12$