Question

How do you graph `3x + 4y < 12` and `x + 3y < 6`, x>0 and y>0?

Answer 1

Graph and solve the system
3x + 4y < 12
x + 3y < 6

Explanation:

Bring these inequalities to standard form:
(1) 3x + 4y - 12 < 0
(2) x + 3y - 6 < 0

First, graph the line: 3x + 4y - 12 = 0 by its 2 intercepts.
Make x = 0 --> y = 3. Make y = 0 --> x = 4.
The solution set of inequality (1) is the area below the line and is limited by the 2 axis Ox and Oy (x > 0 and y > 0)
Next, graph line -> x + 3y - 6 = 0 by its 2 intercepts.
The solution set of inequality (2) is the area below this line and is limited by the 2 axis Ox and Oy (x > 0 and y > 0).
The compound solution set is the commonly shared area.
graph{3x + 4y - 12 = 0 [-10, 10, -5, 5]}
graph{x + 3y - 6 = 0 [-10, 10, -5, 5]}