Write four solutions for the following equation: $2x + 3y = 7$.

(N/A) To find the solutions for the linear equation $2x + 3y = 7$,we can express $y$ in terms of $x$:
$3y = 7 - 2x$
$y = \frac{7 - 2x}{3}$
Now,we substitute different values of $x$ to find corresponding values of $y$:
$1$. If $x = -1$,then $y = \frac{7 - 2(-1)}{3} = \frac{7 + 2}{3} = \frac{9}{3} = 3$. Solution: $(-1, 3)$.
$2$. If $x = 2$,then $y = \frac{7 - 2(2)}{3} = \frac{7 - 4}{3} = \frac{3}{3} = 1$. Solution: $(2, 1)$.
$3$. If $x = 5$,then $y = \frac{7 - 2(5)}{3} = \frac{7 - 10}{3} = \frac{-3}{3} = -1$. Solution: $(5, -1)$.
$4$. If $x = 8$,then $y = \frac{7 - 2(8)}{3} = \frac{7 - 16}{3} = \frac{-9}{3} = -3$. Solution: $(8, -3)$.
Thus,four solutions are $(-1, 3), (2, 1), (5, -1), (8, -3)$.