The range of the values of $x$ which satisfies the inequation $(\frac{3 x + 2}{2 x + 3}) \geq 4$ is

x \in [- 2, - \frac{3}{2})

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x \in (- 2, - \frac{3}{2})

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x \in [- \frac{2}{3}, - \frac{3}{2}]

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x \in [- 2, \frac{3}{2}]

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Solution

The correct option is A $x \in [- 2, - \frac{3}{2})$
$\frac{3 x + 2}{2 x + 3} \geq 4$

$=> \frac{3 x + 2}{2 x + 3} - 4 \geq 0$

$=> \frac{3 x + 2 - 8 x - 12}{2 x + 3} \geq 0$

$\frac{- 5 x - 10}{2 x + 3} \geq 0$

$\frac{5 x + 10}{2 x + 3} \leq 0$
Now, $5 x + 10 \geq 0$ and $2 x + 3 < 0$
=> $x \geq - 2$ and $x < - \frac{3}{2}$
Thus, $x \in [- 2, - \frac{3}{2})$

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