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