Write four solutions for the following equation: $3x + 5y = 0$.

(N/A) Given equation: $3x + 5y = 0$.
Rearranging for $y$:
$5y = -3x$
$y = \frac{-3x}{5}$
To find solutions,we substitute different values for $x$:
$1$. If $x = 0$,then $y = \frac{-3(0)}{5} = 0$. So,$(0, 0)$ is a solution.
$2$. If $x = 5$,then $y = \frac{-3(5)}{5} = -3$. So,$(5, -3)$ is a solution.
$3$. If $x = -5$,then $y = \frac{-3(-5)}{5} = 3$. So,$(-5, 3)$ is a solution.
$4$. If $x = 10$,then $y = \frac{-3(10)}{5} = -6$. So,$(10, -6)$ is a solution.
Thus,four solutions of the given equation are $(0, 0), (5, -3), (-5, 3),$ and $(10, -6)$.