## Precalculus (6th Edition)

$\color{blue}{(-\infty, +\infty)}$
Distribute $3$ to obtain: $3(x) + 3(5)+1 \ge 5+3x \\3x+15+1 \ge 5+3x \\3x+16 \ge 5+3x$ Subtract $16$ and $3x$ to both sides, then combine like terms to obtain: $3x+16-16-3x\ge 5+3x-16-3x \\0\ge -11$ The resulting statement above is true, which means that any real numbers is a solution. Thus , the solution set is: $\color{blue}{(-\infty, +\infty)}$