Question

How do you graph `y < 7/3x + 52/3`?

Answer 1

Draw a dotted straight line graph and shade the area below the graph.

Explanation:

Graphing a linear inequality is the same as graphing a straight line.

In this case, the `y`-intercept is `17 1/3` and the slope is `7/3`

The straight line will be drawn as a dotted line. The shaded area indicating the required region will be BELOW the line.

Use the origin as a point to check:

For `(0,0)`, in the equation `" "y < 1/3x+52/3`

`0 < 51 1/3" "larr` this is true, so the origin lies in the required region.

graph{y< 7/3x+52/3 [-28.54, 51.46, -7.88, 32.12]}