Here are two situations and two equations that represent them.

Here are graphs of the equations. On each graph, the coordinates of some points are shown.

Graph of a line, origin O. Horizontal axis, number of school days, scale is 0 to 20, by 2’s. Vertical axis, bus pass balance in dollars, scale is 0 to 50, by 5’s. Line starts at 0 comma 40, passes through 2 comma 35, 10 comma 15, and 16 comma 0.

Graph of a line, origin O. Horizontal axis, bags of popcorn sold, scale is 0 to 22 by 2’s. Vertical axis, cups of ice tea sold, scale is 0 to 60 by 5’s. Line starts at 0 comma 40, passes through 2 comma 36, 10 comma 20, and 20 comma 0.

The 40 in the first equation can be observed on the graph and the -2.50 can be found with a quick calculation. The graph intersects the vertical axis at 40 and the -2.50 is the slope of the line. Every time increases by 1, decreases by 2.50. In other words, with each passing school day, the dollar amount in Mai's bus pass drops by 2.50. 

The numbers in the second equation are not as apparent on the graph. The values where the line intersects the vertical and horizontal axes, 40 and 20, are not in the equation. We can, however, reason about where they come from.

• If  is 0 (no popcorn is sold), the club would need to sell 40 cups of iced tea to make $60 because .
• If  is 0 (no iced tea is sold), the club would need to sell 20 bags of popcorn to make $60 because . 

What about the slope of the second graph? We can compute it from the graph, but it is not shown in the equation .

Notice that in the first equation, the variable was isolated. Let’s rewrite the second equation and isolate :

Now the numbers in the equation can be more easily related to the graph: The 40 is where the graph intersects the vertical axis and the -2 is the slope. The slope tells us that as increases by 1, falls by 2. In other words, for every additional bag of popcorn sold, the club can sell 2 fewer cups of iced tea.