How do you graph the inequality #3/4x + 1/4y 1#?

Question

How do you graph the inequality #3/4x + 1/4y > 1#?

Julia Adams · Accepted Answer

Draw the straight line #(3x+y)/4=1# that passes through# (4/3, 0) and (0, 4)#. Shade the region above this line in the positive y-direction. Enter therein the given inequality.

The given inequality can be rearranged to the form #y>4-3x# Draw the straight line #y=4-3x# that passes through# (4/3, 0) and (0, 4)#. Shade the region above this line in the positive y-direction. Enter therein the given inequality. For any point (x, y) in the shaded region, #y>4-3x#, and so, #y/4+(3x)/4>1#.

Hazel Anglin · Answer

undefined To graph the inequality $ \frac{3}{4}x + \frac{1}{4}y > 1 $, follow these steps:

1. Rewrite the inequality in slope-intercept form: $ y > -3x + 4 $.
2. Graph the boundary line $ y = -3x + 4 $ as a dashed line because the inequality is strict (>).
3. Choose a test point not on the boundary line, such as the origin (0,0).
4. Substitute the coordinates of the test point into the inequality. If the inequality is true, shade the region containing the test point. If false, shade the opposite region.
5. Since $ 0 > 1 $ is false, shade the region not containing the origin.

This shaded region represents the solution set of the inequality.

How do you graph the inequality #3/4x + 1/4y &gt; 1#?

How do you graph the inequality #3/4x + 1/4y > 1#?