Answer:
[tex]\text{x}\geq \frac{-4(2+\text{a})}{(3-\text{a})}[/tex]
Step-by-step explanation:
Given: the inequality 3x+8+2ax≥3ax−4a
To Find: value of x
Solution:
in given inequality
[tex]3\text{x}+8+2\text{a}\text{x}\geq 3\text{a}\text{x}-4\text{a}[/tex]
[tex]3\text{x}+2\text{a}\text{x}-3\text{a}\text{x}\geq -4\text{a}-8[/tex]
[tex]\text{x}(3+2\text{a}-3\text{a})\geq -8-4\text{a}[/tex]
[tex]\text{x}(3-\text{a})\geq -8-4\text{a}[/tex]
[tex]\text{x}\geq \frac{-8-4\text{a}}{(3-\text{a})}[/tex]
[tex]\text{x}\geq \frac{-4(2+\text{a})}{(3-\text{a})}[/tex]
therefore value of [tex]\text{x}\geq \frac{-4(2+\text{a})}{(3-\text{a})}[/tex]
