## Algebra 2 (1st Edition)

$x=7,y=1$ and $x=3,y=4$.
$x$ and $y$ must both be positive whole numbers. Thus we try the possible values: If $y=0$ then $x=\frac{25-4(0)}{3}=\frac{25}{3}$, this is not a whole number, thus not a solution. If $y=1$ then $x=\frac{25-4(1)}{3}=\frac{21}{3}=7$, this is a whole number, thus a solution. If $y=2$ then $x=\frac{25-4(2)}{3}=\frac{17}{3}$, this is not a whole number, thus not a solution. If $y=3$ then $x=\frac{25-4(3)}{3}=\frac{13}{3}$, this is not a whole number, thus not a solution. If $y=4$ then $x=\frac{25-4(4)}{3}=\frac{9}{3}=3$, this is a whole number, thus a solution. If $y=5$ then $x=\frac{25-4(5)}{3}=\frac{5}{3}$, this is not a whole number, thus not a solution. If $y=6$ then $x=\frac{25-4(6)}{3}=\frac{1}{3}$, this is not a whole number, thus not a solution. For larger values of $y$, $x$ would be negative, so there are no other good solutions. The solutions are $x=7, y=1$ and $x=3,y=4$.