3x + 9 ≥ −x + 19

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 $\left[\frac{5}{2}, \infty\right)$.

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