The given inequlaity is:
[tex]\frac{5(x-3)}{2}-1\leq9[/tex]Solve as follows:
[tex]\begin{gathered} \frac{5(x-3)}{2}-1+1\leq9+1 \\ \frac{5}{2}(x-3)\leq10 \\ \frac{x-3}{2}\leq\frac{10}{5} \\ \frac{x-3}{2}\leq2 \\ x-3\leq4 \\ x\leq7 \end{gathered}[/tex]So the solution is:
[tex]x\leq7[/tex]