Answer :
We begin by translating the statement into an inequality. The phrase “eleven more than three times a number is at most 62” translates to
[tex]$$
3x + 11 \le 62.
$$[/tex]
Step 1: Isolate the term with [tex]\( x \)[/tex].
Subtract 11 from both sides:
[tex]$$
3x + 11 - 11 \le 62 - 11,
$$[/tex]
which simplifies to
[tex]$$
3x \le 51.
$$[/tex]
Step 2: Solve for [tex]\( x \)[/tex].
Divide both sides of the inequality by 3:
[tex]$$
\dfrac{3x}{3} \le \dfrac{51}{3},
$$[/tex]
yielding
[tex]$$
x \le 17.
$$[/tex]
Thus, the solution to the inequality is:
[tex]$$
\boxed{x \le 17}.
$$[/tex]
[tex]$$
3x + 11 \le 62.
$$[/tex]
Step 1: Isolate the term with [tex]\( x \)[/tex].
Subtract 11 from both sides:
[tex]$$
3x + 11 - 11 \le 62 - 11,
$$[/tex]
which simplifies to
[tex]$$
3x \le 51.
$$[/tex]
Step 2: Solve for [tex]\( x \)[/tex].
Divide both sides of the inequality by 3:
[tex]$$
\dfrac{3x}{3} \le \dfrac{51}{3},
$$[/tex]
yielding
[tex]$$
x \le 17.
$$[/tex]
Thus, the solution to the inequality is:
[tex]$$
\boxed{x \le 17}.
$$[/tex]
