If 4x - 3y = 12 and xy = 5 , then find the value of$\frac{16x^2+9y^2}{8}$

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If 4x - 3y = 12 and xy = 5 , then find the value of $\frac{16x^2+9y^2}{8}$

A33

B15

C18

D44

Answer:

4x - 3y = 12

$(4x-3y)^2=12^2$

$16x^2+9y^2-24xy=144$

$16x^2+9y^2-24\times5 = 144$

$16x^2+9y^2=144+120=264$

$\frac{16x^2+9y^2}{8}=264/8$

$=33$

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