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

##### 1 Answer
Dec 5, 2017

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 $17 \frac{1}{3}$ and the slope is $\frac{7}{3}$

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 $\left(0 , 0\right)$, in the equation $\text{ } y < \frac{1}{3} x + \frac{52}{3}$

$0 < 51 \frac{1}{3} \text{ } \leftarrow$ 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]}