How do you graph the system #-3x+2y3# and #x+4y

Question

How do you graph the system #-3x+2y>3# and #x+4y<-2#?

Isaiah Anthony · Accepted Answer

Solve the system
(1) -3x + 2y > 3
(2) x + 4y < -2

Firstly, standardize the two inequalities as follows: (1) -3x + 2y - 3 > 0; (2) x + 4y + 2 < 0 Utilizing the test point origin O(0, 0), graph the line (1): f(x, y) = -3x + 2y - 3 by its two intercepts. Set x = 0 --> y = 3/2. Set y = 0, --> x = -1. We obtain -3 > 0 - which is untrue. The region that is free of O is the solution set; color or shade it. After that, graph Line (2): f(x, y) = x + 4y - 2 by its two intercepts. Let's say that x = 0 --> y = -1/2. Let's say that y = 0, --> x = - 2 Test point origin O(0, 0). Then, the area that does not contain O is the solution set; shade or color it. The area that is shared by both solution sets is the combined solution set.

Avery Alexander · Answer

undefined To graph the system of inequalities $-3x + 2y > 3$ and $x + 4y < -2$, you would first graph the boundary lines for each inequality separately. For $-3x + 2y > 3$, you would graph the boundary line $ -3x + 2y = 3 $ as a dashed line (since it's a greater than inequality). For $x + 4y < -2$, you would graph the boundary line $x + 4y = -2$ as a dashed line (since it's a less than inequality). Then, you would shade the regions that satisfy both inequalities.

How do you graph the system #-3x+2y&gt;3# and #x+4y&lt;-2#?

How do you graph the system #-3x+2y>3# and #x+4y<-2#?