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Question
3x + 9 ≥ −x + 19
Solution
\[3x + 9 \geqslant - x + 19\]
\[ \Rightarrow 3x + x \geqslant 19 - 9 (\text{ Transposing - x to the LHS } \hspace{0.167em} and 9 to the RHS)\]
\[ \Rightarrow 4x \geq 10\]
\[ \Rightarrow x \geq \frac{5}{2} (\text{ Dividing both the sides by 4) }\]
\[\text{ Hence, the solution set of the given inequation is } [\frac{5}{2}, \infty ) .\]
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