we have
[tex]4(x + 3) \leq 0[/tex] -------> inequality 1
or
[tex]x + 1 > 3[/tex] -------> inequality 2
we know that
In this system of inequalities, for a value to be the solution of the system, it is enough that it satisfies at least one of the two inequalities.
let's check each of the values
case 1) x=-6
Substitute the value of x=-6 in the inequality 1
[tex]4(-6 + 3) \leq 0[/tex]
[tex]4(-3) \leq 0[/tex]
[tex]-12 \leq 0[/tex] -------> is ok
The value of x=-6 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies
case 2) x=-3
Substitute the value of x=-3 in the inequality 1
[tex]4(-3 + 3) \leq 0[/tex]
[tex]4(0) \leq 0[/tex]
[tex]0 \leq 0[/tex] -------> is ok
The value of x=-3 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies
case 3) x=0
Substitute the value of x=0 in the inequality 1
[tex]4(0 + 3) \leq 0[/tex]
[tex]4(3) \leq 0[/tex]
[tex]12 \leq 0[/tex] -------> is not ok
Substitute the value of x=0 in the inequality 2
[tex]0 + 1 > 3[/tex]
[tex] 1 > 3[/tex] --------> is not ok
The value of x=0 is not a solution of the compound inequality
case 4) x=3
Substitute the value of x=3 in the inequality 1
[tex]4(3 + 3) \leq 0[/tex]
[tex]4(6) \leq 0[/tex]
[tex]24 \leq 0[/tex] -------> is not ok
Substitute the value of x=3 in the inequality 2
[tex]3 + 1 > 3[/tex]
[tex] 4 > 3[/tex] --------> is ok
The value of x=3 is a solution of the compound inequality
case 5) x=8
Substitute the value of x=8 in the inequality 1
[tex]4(8 + 3) \leq 0[/tex]
[tex]4(11) \leq 0[/tex]
[tex]44 \leq 0[/tex] -------> is not ok
Substitute the value of x=8 in the inequality 2
[tex]8 + 1 > 3[/tex]
[tex] 9 > 3[/tex] --------> is ok
The value of x=8 is a solution of the compound inequality
case 6) x=10
Substitute the value of x=10 in the inequality 1
[tex]4(10 + 3) \leq 0[/tex]
[tex]4(13) \leq 0[/tex]
[tex]52 \leq 0[/tex] -------> is not ok
Substitute the value of x=10 in the inequality 2
[tex]10+ 1 > 3[/tex]
[tex] 11 > 3[/tex] --------> is ok
The value of x=10 is a solution of the compound inequality
therefore
the answer is
[-6,-3,3,8,10]
