\( 6 – 2x ≥ \dfrac{x}{5} + 4\)
\(\dfrac{ 6 – 2x ≥ (x+20)}{5}\)
5(6 – 2x) ≥ (x + 20)
30 – 10x ≥ x + 20
30 – 20 ≥ x + 10x
10 ≥11x
11x ≤ 10
Divide both sides by 11, we get
\(\dfrac{ 11x}{11} ≤ \dfrac{10}{11}\)
x ≤ \( \dfrac{10}{11}\)
The solution of the given inequation is (-∞, \( \dfrac{10}{11}\)).
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