Show that the points $A(1, 2)$,$B(-1, -16)$,and $C(0, -7)$ lie on the graph of the linear equation $y = 9x - 7$.

(N/A) To show that a point $(x, y)$ lies on the graph of the linear equation $y = 9x - 7$,we must substitute the coordinates of the point into the equation and verify if the left-hand side equals the right-hand side.
For point $A(1, 2)$:
Substitute $x = 1$ and $y = 2$ into the equation:
$2 = 9(1) - 7$
$2 = 9 - 7$
$2 = 2$
Since the left-hand side equals the right-hand side,point $A$ lies on the graph.
For point $B(-1, -16)$:
Substitute $x = -1$ and $y = -16$ into the equation:
$-16 = 9(-1) - 7$
$-16 = -9 - 7$
$-16 = -16$
Since the left-hand side equals the right-hand side,point $B$ lies on the graph.
For point $C(0, -7)$:
Substitute $x = 0$ and $y = -7$ into the equation:
$-7 = 9(0) - 7$
$-7 = 0 - 7$
$-7 = -7$
Since the left-hand side equals the right-hand side,point $C$ lies on the graph.
Thus,all three points $A$,$B$,and $C$ satisfy the equation $y = 9x - 7$ and therefore lie on its graph.