Solve the following


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


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

$\Rightarrow \frac{4+2 x}{3}-\frac{x}{2} \geq-3$

$\Rightarrow \frac{8+4 x-3 x}{6} \geq-3$

$\Rightarrow 8+x \geq-18 \quad$ [Multiplying both the sides by 6$]$

$\Rightarrow x \geq-26 \quad$ [Transposing 8 to the RHS]

Thus, the solution set of the given inequation is $[-26, \infty)$.

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