$$2 (3 – x) ≥ \dfrac{x}{5} + 4$$

Asked by Aaryan | 1 year ago |  149

##### Solution :-

$$6 – 2x ≥ \dfrac{x}{5} + 4$$

$$\dfrac{ 6 – 2x ≥ (x+20)}{5}$$

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 (-∞, $$\dfrac{10}{11}$$).

Answered by Aaryan | 1 year ago

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