Note: Options are missing. So, the general solution of the given inequality is shown below.
Given:
The inequality is:
[tex]\dfrac{1}{2}(x+2)+3y<8[/tex]
To find:
The point lies in the solution set of the given inequality.
Solution:
We have,
[tex]\dfrac{1}{2}(x+2)+3y<8[/tex]
It can be written as:
[tex]\dfrac{1}{2}(x)+\dfrac{1}{2}(2)+3y<8[/tex]
[tex]\dfrac{1}{2}x+1+3y<8[/tex]
[tex]\dfrac{1}{2}x+3y<8-1[/tex]
[tex]\dfrac{1}{2}x+3y<7[/tex]
All the points that satisfy the above inequality are in the solution set of the given inequality.
For example (0,0).
[tex]\dfrac{1}{2}(0)+3(0)<7[/tex]
[tex]0<7[/tex]
This statement is true. It means (0,0) is in the solution set.
For example (0,3).
[tex]\dfrac{1}{2}(0)+3(3)<7[/tex]
[tex]9<7[/tex]
This statement is false. It means (0,3) is not in the solution set.
The graph of the inequality [tex]\dfrac{1}{2}x+3y<7[/tex] is shown below.
All the points in the shaded region are in the solution set but the points on the boundary line are not in the solution set.
