If 3x - 7y = 10 and xy = -1,


If $3 x-7 y=10$ and $x y=-1$, find the value of $9 x^{2}+49 y^{2}$.


We have,

$(2-7 y)^{2}=(3 x)^{2}+(-7 y)^{2}-2 * 3 x * 7 y$

$\Rightarrow(3 x-7 y)^{2}=9 x^{2}+49 y^{2}-42 x y \quad[$ Since, $3 x-7 y=10$ and $x y=-1]$

$\Rightarrow(10)^{2}=9 x^{2}+49 y^{2}+42$

$\Rightarrow 100-42=9 x^{2}+49 y^{2}$

$\Rightarrow 9 x^{2}+49 y^{2}=58$

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