How do you graph the system #y<= 2x + 3# and #x > 1#?

1 Answer
Jul 23, 2015

Draw the lines (initially dashed) for #y=2x+3# and for #x=1#.
Shade that section that included by inequalities.
Convert the dashed line along this section corresponding to #y=2x+3# into a solid line.

Explanation:

Step 1: draw dashed line for #y = 2x+3#
Evaluate #y = 2x+3# for a couple of values of #x#,
for example #x=0 rarr y= 3# and #x=2 rarr y=7#
Draw the (dashed) line through the corresponding points on the Cartesian plane.

Step 2: draw dashed line for #x=1#
Draw a vertical (dashed) line through #x=1#.

Step 3: Identify the section included in both inequalities
Test a few #(x,y)# pair values until you identify a pair that is valid for both inequalities.
For example #(x,y) = (2, 0)# satisfies both inequalities, since
#color(white)("XXXX")##(0) < 2(2) +3# and # (2) > 1#

Step 4: Shade the section which includes the point that satisfies both inequalities

**Step 5: Convert the dashed line segment bordering the shaded area that corresponds to #y = 2x+3# into a solid line to show that it is included in the relation #y <= 2x +3#

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