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

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Answer 1 · 2017-12-05T18:49:16

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

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]}