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

Jul 26, 2015

Solve system of 3 linear inequalities in 2 variables:
$x + y \ge 2$
$8 x - 2 y \le 16$
$4 y \le 6 x + 8$

Explanation:

Bring all inequalities to standard form
(1) $x + y - 2 \ge 0$
(2) $8 x - 2 y - 16 \le 0$
(3) $4 y - 6 x - 8 \le 0$
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:$- 2 \ge 0$. 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: $- 16 \le 0.$ 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:$- 8 \le 0$. 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) $2 + 3 - 2 \ge 0$ OK
(2) $16 - 6 - 16 \le 0$ OK
(3) $12 - 12 - 8 \le 0$ OK
NOTE. The 3 lines are included in the solution set because the presence of the sign $\le$ or $\ge$