#### Answer

$-1\le x \le \dfrac{1}{5}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To find all $x$ such that $f(x)
\le3
,$ with $f(x)=
|5x+2|
,$ solve the inequality
\begin{array}{l}\require{cancel}
|5x+2|\le3
.\end{array}
Then graph the solution set.
In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$
$\bf{\text{Solution Details:}}$
Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-3\le 5x+2 \le3
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-3\le 5x+2 \le3
\\\\
-3-2\le 5x+2-2 \le3-2
\\\\
-5\le 5x \le 1
\\\\
-\dfrac{5}{5}\le \dfrac{5x}{5} \le \dfrac{1}{5}
\\\\
-1\le x \le \dfrac{1}{5}
.\end{array}
Hence, the solution set $
-1\le x \le \dfrac{1}{5}
.$