Answer:
[tex]\textsf{A.} \quad \left(- \infty, -\dfrac{11}{3}\right][/tex]
Step-by-step explanation:
Given inequality:
[tex]\dfrac{6x+2}{5}-\dfrac{3x-7}{9} \leq -2[/tex]
Make the denominators of both fractions the same:
[tex]\implies \dfrac{9(6x+2)}{9 \cdot 5}-\dfrac{5(3x-7)}{9 \cdot 5} \leq -2[/tex]
[tex]\implies \dfrac{54x+18}{45}-\dfrac{15x-35}{45} \leq -2[/tex]
Combine the fractions:
[tex]\implies \dfrac{54x+18-(15x-35)}{45} \leq -2[/tex]
[tex]\implies \dfrac{54x+18-15x+35}{45} \leq -2[/tex]
[tex]\implies \dfrac{39x+53}{45} \leq -2[/tex]
Multiply both sides by 45:
[tex]\implies 39x+53\leq -90[/tex]
Subtract 53 from both sides:
[tex]\implies 39x\leq -143[/tex]
Divide both sides by 39:
[tex]\implies x\leq -\dfrac{11}{3}[/tex]
Therefore, the solution set is:
[tex]\left(- \infty, -\dfrac{11}{3}\right][/tex]
