Answer:
(a)The solution set is: [tex]x \in [6, \infty) \forall x \in R[/tex]
(c) [tex]6 \leq x[/tex]
Step-by-step explanation:
Given the inequality: [tex]12+\dfrac{11}{6}x\leq 5+3x[/tex]
We solve by collecting like terms
[tex]12+\dfrac{11}{6}x\leq 5+3x\\12-5\leq 3x-\dfrac{11}{6}x\\7\leq \dfrac{18x-11x}{6}\\42 \leq 7x\\$Divide both sides by 7\\6 \leq x\\$We can re-write this as:\\x\geq 6[/tex]
The solution set is therefore: [tex]x \in [6, \infty) \forall x \in R[/tex]
