Answer:
[tex]a=3,\text{ then } x\in(-\infty,\infty)\\ \\a\in(-\infty,3),\text{ then }x\in \left[\dfrac{-4a-8}{3-a},\infty\right)\\ \\a\in(3,\infty),\text{ then }x\in \left(-\infty, \dfrac{-4a-8}{3-a}\right][/tex]
Step-by-step explanation:
In the inequality
[tex]3x+8+2ax\ge 3ax-4a\ \ \ (1)[/tex]
a is an arbitrary real number.
Separate the terms with x into left side and the terms without x in the right side:
[tex]3x+2ax-3ax\ge -4a-8\\ \\3x-ax\ge -4a-8\\ \\(3-a)x\ge -4a-8\ \ \ (2)[/tex]
First, look at the leading coefficient at x. If this coefficient is equal to 0 (when [tex]a=3[/tex]), then the inequality is
[tex]0\ge -4a-8\\ \\0\ge -12-8\ [\text{Substituted }a=3]\\ \\0\ge -20[/tex]
This is true inequality for all x, so at [tex]a=3,[/tex] the inequality (1) has the solution [tex]x\in (-\infty,\infty)[/tex]
Now, if the leading coefficient
[tex]3-a>0\\ \\a<3,[/tex]
then we can divide the inequality (2) by this positive number [tex]3-a[/tex] and get
[tex]x\ge \dfrac{-4a-8}{3-a}[/tex]
If the leading coefficient
[tex]3-a<0\\ \\a>3,[/tex]
then we can divide the inequality (2) by this negative number [tex]3-a[/tex] and get
[tex]x\le \dfrac{-4a-8}{3-a}[/tex]
So, the answer is
[tex]a=3,\text{ then } x\in(-\infty,\infty)\\ \\a\in(-\infty,3),\text{ then }x\in \left[\dfrac{-4a-8}{3-a},\infty\right)\\ \\a\in(3,\infty),\text{ then }x\in \left(-\infty, \dfrac{-4a-8}{3-a}\right][/tex]
