Question:
If $9 x^{2}+25 y^{2}=181$ and $x y=-6$, find the value of $3 x+5 y$.
Solution:
We have,
$(3 x+5 y)^{2}=(3 x)^{2}+(5 y)^{2}+2 * 3 x * 5 y$
$\Rightarrow(3 x+5 y)^{2}=9 x^{2}+25 y^{2}+30 x y$
$=181+30(-6) \quad\left[\right.$ Since, $9 x^{2}+25 y^{2}=181$ and $\left.x y=-6\right]$
$\Rightarrow(3 x+5 y)^{2}=1$
$\Rightarrow(3 x+5 y)^{2}=(\pm 1)^{2}$
$\Rightarrow 3 x+5 y=\pm 1$