Question

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Answer 1

$= >( {3x+ 2y}^{2} ) = {7}^{2} \\ = > 9 {x}^{2} + 12xy + 4 {y}^{2} = 49 \\ = > 9 {x}^{2} + 12 \times 11 + 4 {y}^{2} = 49 \\ = > 9 {x}^{2} + 132 + 4 {y}^{2} = 49 \: (xy = 11) \\ = > 9 {x}^{2} + 4 {y}^{2} = 49 - 132 \\ = > 9 {x}^{2} + 4 {y}^{2} = - 83$
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Answer 2

OK LET'S SEE AS I AM THE GREATEST

SOLUTION:

3X+2Y=7.......EQ1

XY=11

SQUARING BOTH SIDES OF EQ1

USING IDENTITY

(A+B)^2=A^2+B^2 +2AB

WE GET

9X^2+4Y^2+12XY=49

9X^2+4Y^2=49-12(11)

9X^2+4Y^2=49-132

9X^2+4Y^2= -83

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