If $4 x^{2} + 9 y^{2} = 40, x y = 2$ , find $2 x + 3 y$

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Solution

$(2 x)^{2} + (3 y)^{2} = 40$
$(2 x) (3 y) = 12$
$(2 x + 3 y)^{2} = (2 x)^{2} + (3 y)^{2} + 2 (2 x) (3 y)$
$= 40 + 24$
$= 64$
$2 x + 3 y = \pm 8$

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