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Q.Solve the inequality : 12(3x5+4)≥13(x−6)\dfrac{1}{2}\left(\dfrac{3x}{5} + 4\right) \geq \dfrac{1}{3}(x - 6).

Jammu Kashmir JkboseJammu and Kashmir Board of School Education (Class 11) 2025Subjective· 4mImportance★★★★★
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Expanding and simplifying 12(3x5+4)≥13(x−6)\frac12(\frac{3x}{5}+4)\ge\frac13(x-6) leads to x≤120x\le120.

Expand the left side:

12(3x5+4)=3x10+2.\dfrac12\left(\dfrac{3x}{5}+4\right) = \dfrac{3x}{10} + 2.

Expand the right side:

13(x−6)=x3−2.\dfrac13(x-6) = \dfrac{x}{3} - 2.

The inequality becomes:

3x10+2≥x3−2.\dfrac{3x}{10} + 2 \ge \dfrac{x}{3} - 2.

Bring the xx-terms to one side and constants to the other:

3x10−x3≥−2−2=−4.\dfrac{3x}{10} - \dfrac{x}{3} \ge -2-2 = -4.

Combine the fractions using LCM 30: …

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