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semsee45

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]

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