Answer:
(3,-8)
(2,-10)
Step-by-step explanation:
we have
[tex]10x + 4y < 12[/tex] ----> inequality A
[tex]8x- 3y > 20[/tex] ----> inequality B
we know that
If a point is a solution of the system of inequalities, then the point must satisfy both inequalities (makes true both inequalities)
Verify all the points
Substitute the value of x and the value of y of each point in both inequalities
Case 1) point (3,-8)
For x=3, y=-8
inequality A
[tex]10(3) + 4(-8) < 12[/tex]
[tex]-2 < 12[/tex] ----> is true
inequality B
[tex]8(3)- 3(-8) > 20[/tex]
[tex]48 > 20[/tex] ---> is true
therefore
The point is a solution of the system of linear inequalities
Case 2) point (2,5)
For x=2, y=5
inequality A
[tex]10(2) + 4(5) < 12[/tex]
[tex]40 < 12[/tex] ----> is not true
therefore
The point is not a solution of the system of linear inequalities
Case 3) point (-5,1)
For x=-5, y=1
inequality A
[tex]10(-5) + 4(1) < 12[/tex]
[tex]-46 < 12[/tex] ----> is true
inequality B
[tex]8(-5)- 3(1) > 20[/tex]
[tex]-43 > 20[/tex] ---> is not true
therefore
The point is not a solution of the system of linear inequalities
Case 4) point (10,3)
For x=10, y=3
inequality A
[tex]10(10) + 4(3) < 12[/tex]
[tex]112 < 12[/tex] ----> is not true
therefore
The point is not a solution of the system of linear inequalities
Case 5) point (2,-10)
For x=2, y=-10
inequality A
[tex]10(2) + 4(-10) < 12[/tex]
[tex]-20 < 12[/tex] ----> is true
inequality B
[tex]8(2)- 3(-10) > 20[/tex]
[tex]46 > 20[/tex] ---> is true
therefore
The point is a solution of the system of linear inequalities
see the attached figure to better understand the problem
If a ordered pair is a solution of the system , then the ordered pair must lie in the shaded area of the solution set
