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(A) Given inequality: $3(2x - 1) + 5 \leq \frac{1}{2}(x + 15)$ Multiply both sides by $2$ to clear the fraction: $6(2x - 1) + 10 \leq x + 15$ Expand t

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$x \in (-\infty, 1]$

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(A) Given inequality: $3(2x - 1) + 5 \leq \frac{1}{2}(x + 15)$ Multiply both sides by $2$ to clear the fraction: $6(2x - 1) + 10 \leq x + 15$ Expand the terms: $12x - 6 + 10 \leq x + 15$ Simplify: $12x + 4 \leq x + 15$ Subtract $x$ from both sides: $11x + 4 \leq 15$ Subtract $4$ from both sides: $11x \leq 11$ Divide by $11$: $x \leq 1$ Thus,the solution set is $x \in (-\infty, 1]$.

Solve the inequality for real $x$: $3(2x - 1) + 5 \leq \frac{1}{2}(x + 15)$

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