Question:
3x + 9 ≥ −x + 19
Solution:
$3 x+9 \geqslant-x+19$
$\Rightarrow 3 x+x \geqslant 19-9$ (Transposing $-x$ to the LHS and 9 to the RHS)
$\Rightarrow 4 x \geq 10$
$\Rightarrow x \geq \frac{5}{2}$ (Dividing both the sides by 4 )
Hence, the solution set of the given inequation is $\left[\frac{5}{2}, \infty\right)$.