How do you graph #2x – 4y > 8# and #y <3x + 1#?

to get the intersections with the coordinate axes for with the x axis y=0

now with the y axis, x=0 when x=0

then both points are located and the straight line is obtained by being an inequality and having the greater sign, a line segment and shaded down is graphed

same for y=0

for x=0

Graph the region above the line (2x - 4y = 8) (dashed line) and below the line (y = 3x + 1) (solid line). Shade the region that satisfies both inequalities.

To graph the system of inequalities (2x - 4y > 8) and (y < 3x + 1), follow these steps:

1. Graph the boundary line for each inequality.
2. Decide whether the line should be solid or dashed based on the inequality sign.
3. Determine which side of each line represents the solution region.
4. Shade the appropriate regions.
5. Identify the overlapping shaded region, if any.

For the inequality (2x - 4y > 8):

• Graph the line (2x - 4y = 8).
• Since it's (>), use a dashed line.
• Decide which side of the line to shade. You can choose a point not on the line (0,0) is convenient.
• Shade the region where the inequality holds true.

For the inequality (y < 3x + 1):

• Graph the line (y = 3x + 1).
• Since it's (<), use a dashed line.
• Decide which side of the line to shade. Again, you can use a point not on the line.
• Shade the region where the inequality holds true.

The shaded regions represent where both inequalities are satisfied. If there's any overlapping shaded region, that's the solution region for the system of inequalities.