Answer:
[tex]\boxed{18. \, -3 \leq x \leq \dfrac{16}{9};\quad 22. \text{ $x \leq -19$ or $x \geq -5$}}[/tex]
Step-by-step explanation:
Q18
[tex]\left |\dfrac{3x - 5}{2} \right | \leq 7[/tex]
Multiply each side by 2
[tex]|3x - 5| \leq 14[/tex]
Apply the absolute rule:
If |x| ≤ a, then -a ≤ x ≤ a
This gives
-14 ≤ 3x - 5 ≤ 14
Separate into two inequalities
[tex]\begin{array}{cc}-14 \leq 3x - 5 & 3x - 5 \leq 14 \\-9 \leq 3x & 3x\leq 19 \\-3 \leq x & x \leq \dfrac{19}{3} \\\end{array}[/tex]
Merge the overlapping intervals
[tex]\boxed{\mathbf{-3 \leq x \leq \dfrac{16}{9}}}[/tex]
Q22
-5|x + 12| ≤ -35
Divide each side by -5
This reverses the sign of the inequality.
|x + 12| ≥ 7
Apply the absolute rule
If |x| ≥ a, then -a ≥ x or x ≥ a
-7 ≥ x + 12 or x + 12 ≥ 7
Separate into two inequalities
[tex]\begin{array}{cc}-7 \geq x + 12 & x + 12 \geq 7 \\-19 \geq x & x\geq -5 \\\end{array}[/tex]
Combine the intervals
[tex]\boxed{\textbf{$x \leq -19$ or $x \geq -5$}}[/tex]
