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Answer 1

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Answer 2

Given :-

$\frac{2x - 3}{4} + 9≥ 3 + \frac{4x}{3}$

To Find :-

The value of "x" .

Solution :-

$\frac{2x - 3}{4} + 9≥3 + \frac{4x}{3}$

Separating LCM of LHS we get ,

$\frac{2x}{4} - \frac{3}{4} + \frac{9}{1} ≥ \frac{3}{1} + \frac{4x}{3}$

$\frac{x}{2} - \frac{3}{4} + \frac{9}{1} - \frac{3}{1} - \frac{4x}{ 3} ≥0$

$\frac{6x - 9 + 108 - 36 - 16x}{12}≥0$

$\frac{ - 10x + 63}{12} ≥0$

Again Separating LCM we have ,

$\frac{ - 10x}{12} + \frac{63}{12}≥0$

$\frac{ - 5x}{6} + \frac{21}{4} ≥0$

$\frac{5x}{6} ≤ \frac{21}{4}$

$x ≤ \frac{21 \times 6}{4 \times 5}$

$x≤ \frac{63}{10}$

$x≤6.3$

Henceforth , x ≤ 6.3