Answer:
[tex]-3 \le h \le 1[/tex].
Step-by-step explanation:
Apply the property of absolute values: if [tex]a \ge 0[/tex], then [tex]|x| \le a \iff -a \le x \le a[/tex]. By this property, [tex]|- 3\, h - 6 | \le 3[/tex] is equivalent to [tex]-3 \le -3\, h - 6\le 3[/tex]. That's the same as saying that [tex]-3\, h - 6 \ge -3[/tex] and [tex]-3\, h - 6 \le 3[/tex].
Add [tex]6[/tex] to both sides of both inequalities:
[tex]-3\, h \ge 3[/tex] and [tex]-3\, h \le 9[/tex].
Divide both sides of both inequalities by [tex](-3)[/tex]. Note that because [tex]-3 < 0[/tex], dividing both sides of an equality by this number will flip the direction of the inequality sign.
- [tex]-3\, h \ge 3[/tex] would become [tex]h \le -1[/tex].
- [tex]-3\, h \le 9[/tex] would become [tex]h \ge -3[/tex].
Both inequalities are supposed to be true. Combining the two inequalities to obtain:
[tex]-3 \le h \le 1[/tex].
