The given inequality is
[tex]|5-2x|+1\leq10[/tex]
First, we subtract 1 on each side
[tex]\begin{gathered} |5-2x|+1-1\leq10-1 \\ |5-2x|\leq9 \end{gathered}[/tex]
Now we use the property of inequalities with absolute values, which states
[tex]|x|\leq a\rightarrow-a\leq x\leq a[/tex]
Using this property, we have
[tex]|5-2x|\leq9\rightarrow-9\leq5-2x\leq9[/tex]
We solve the compound inequality now
[tex]\begin{gathered} -9\leq5-2x\leq9 \\ -9-5\leq-2x\leq9-5 \\ -14\leq-2x\leq4 \\ \frac{-14}{-2}\ge\frac{-2x}{-2}\ge\frac{4}{-2} \\ 7\ge x\ge-2 \\ -2\leq x\leq7 \end{gathered}[/tex]
Therefore, the solution is the interval [-2,7], and the graph of it would be
