To practice solving absolute value inequalities that use a 'less than or equal to' symbol, evaluate the expressions $|2x - 1| \leq 5$ and $|4x - 5| \leq 3$. For the first inequality, $|2x - 1| \leq 5$, rewrite it as the equivalent compound inequality $-5 \leq 2x - 1 \leq 5$. Adding $1$ to all parts yields $-4 \leq 2x \leq 6$, and dividing by $2$ gives the solution $-2 \leq x \leq 3$. This is graphed with closed endpoints at $-2$ and $3$ and shading between them, written in interval notation as $[-2, 3]$. For the second inequality, $|4x - 5| \leq 3$, form the compound inequality $-3 \leq 4x - 5 \leq 3$. Adding $5$ across the parts produces $2 \leq 4x \leq 8$, and dividing by $4$ results in $\frac{1}{2} \leq x \leq 2$. Graphed with closed circles and shading in between, the interval notation for this solution is $[\frac{1}{2}, 2]$.