How do you graph #x + y = 2# and #8x - 2y

Question

How do you graph #x + y >= 2# and #8x - 2y <= 16# and #4y <= 6x + 8#?

Eva Adamson · Accepted Answer

Solve system of 3 linear inequalities in 2 variables:
#x + y >= 2#
#8x - 2y <= 16#
#4y <= 6x + 8# Bring all inequalities to standard form (1) #x + y - 2 >= 0# (2) #8x - 2y - 16 <= 0# (3) #4y - 6x - 8 <= 0# First, graph Line (1) -> x + y - 2 = 0 by its 2 intercepts. Make x = 0 --> y = 2. Make y = 0 --> x = 2. Use the origin O as test point. Replace x = 0, y = 0 into (1). We get:# -2 >= 0#. Not true. Then, the solution set area doesn't contain O. Color it. Next, graph Line (2) by its 2 intercepts. Use O as test point. Replace x = 0 and y = 0 into (2). We get: #-16 <= 0.# True. Then the solution set area contains O. Color it. Next, graph Line (3) by its 2 intercepts. Replace x = 0 and y = 0 into (3). We get:# -8 <= 0#. True. Then, the solution set area contains O. The compound solution set is the triangle area, commonly shared by the 3 solution sets. graph{x + y - 2 = 0 [-10, 10, -5, 5]}

Hazel Anglin · Answer

undefined To graph the inequalities $x + y \geq 2$, $8x - 2y \leq 16$, and $4y \leq 6x + 8$:

1. Graph the boundary lines for each inequality:
   - For $x + y \geq 2$, graph the line $x + y = 2$. This line is solid because it includes the "or equal to" component of the inequality.
   - For $8x - 2y \leq 16$, graph the line $8x - 2y = 16$. This line is solid because it includes the "or equal to" component of the inequality.
   - For $4y \leq 6x + 8$, graph the line $4y = 6x + 8$. This line is solid because it includes the "or equal to" component of the inequality.

2. Determine which side of each boundary line to shade by choosing a test point not on the boundary line and substituting its coordinates into the inequality. If the inequality is true, shade the side containing the test point; otherwise, shade the opposite side.

3. Shade the appropriate regions for each inequality:
   - For $x + y \geq 2$, shade the region above the line $x + y = 2$ because the test point (0,0) satisfies the inequality.
   - For $8x - 2y \leq 16$, shade the region below the line $8x - 2y = 16$ because the test point (0,0) satisfies the inequality.
   - For $4y \leq 6x + 8$, shade the region below the line $4y = 6x + 8$ because the test point (0,0) satisfies the inequality.

4. The solution to the system of inequalities is the region where all shaded regions overlap.

5. Optionally, use a different color or pattern for each shaded region to distinguish them on the graph.

By following these steps, you can graph the inequalities $x + y \geq 2$, $8x - 2y \leq 16$, and $4y \leq 6x + 8$ on the coordinate plane.

How do you graph #x + y &gt;= 2# and #8x - 2y &lt;= 16# and #4y &lt;= 6x + 8#?

How do you graph #x + y >= 2# and #8x - 2y <= 16# and #4y <= 6x + 8#?