Question:
$x-2 \leq \frac{5 x+8}{3}$
Solution:
$x-2 \leq \frac{5 x+8}{3}$
$\Rightarrow 3(x-2) \leq 5 x+8 \quad$ [Multiplying both the sides by 3$]$
$\Rightarrow 3 x-6 \leq 5 x+8$
$\Rightarrow 5 x+8 \geq 3 x-6$
$\Rightarrow 5 x-3 x \geq-6-8 \quad$ [Transposing $3 x$ to the LHS and 8 to the RHS]
$\Rightarrow 2 x \geq-14$
$\Rightarrow x \geq-7 \quad$ [Dividing both the sides by 2]
Hence, the solution of the given inequation is $[-7, \infty)$.
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