Question

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

Answer 1

Solve system of 3 linear inequalities in 2 variables:
`x + y >= 2`
`8x - 2y <= 16`
`4y <= 6x + 8`

Explanation:

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]}
graph{8x - 2y - 16 = 0 [-10, 10, -5, 5]}
graph{4y - 6x - 8 = 0 [-10, 10, -5, 5]}
Check with point (2, 3) inside the triangle.
(1) `2 + 3 - 2 >= 0` OK
(2) `16 - 6 - 16 <= 0` OK
(3) `12 - 12 - 8 <= 0` OK
NOTE. The 3 lines are included in the solution set because the presence of the sign `<=` or `>=`