Modulus Inequalities (Absolute Value Inequalities): Problems with Solutions

Solution:
For [tex]c\gt 0[/tex]: [tex]|A|\leq c \Leftrightarrow -c\leq A\leq c[/tex].
[tex]-3\leq x-1\leq 3 \Leftrightarrow -2\leq x\leq 4[/tex]
Geometrically: the distance from [tex]x[/tex] to [tex]1[/tex] on the number line is at most [tex]3[/tex].
Answer: [tex]x\in[-2, 4][/tex].

Solution:
For [tex]c\gt 0[/tex]: [tex]|A|\gt c \Leftrightarrow A\gt c[/tex] or [tex]A\lt -c[/tex].
[tex]2x-1\gt 2 \Rightarrow x\gt \frac{3}{2}[/tex]
[tex]2x-1\lt -2 \Rightarrow x\lt -\frac{1}{2}[/tex]
Answer: [tex]x\in\left(-\infty, -\frac{1}{2}\right)\cup\left(\frac{3}{2}, +\infty\right)[/tex].

Solution:
[tex]|4x-20|=4|x-5|[/tex], so [tex]3|x-5|-4|x-5|=-|x-5|\leq -2[/tex].
[tex]|x-5|\geq 2 \Leftrightarrow x-5\leq -2[/tex] or [tex]x-5\geq 2[/tex]
Answer: [tex]x\in(-\infty, 3]\cup[7, +\infty)[/tex].

Solution:
[tex]x^2+1\gt 0[/tex] for every [tex]x[/tex], so [tex]\left|x^2+1\right|=x^2+1[/tex].
[tex]x^2+1+2x\lt 4 \Leftrightarrow x^2+2x-3\lt 0 \Leftrightarrow (x+3)(x-1)\lt 0[/tex]
Answer: [tex]x\in(-3, 1)[/tex].

Solution:
Domain: [tex]x\neq 2[/tex].
[tex]\frac{1}{|x-2|}\lt 3 \Leftrightarrow |x-2|\gt \frac{1}{3}[/tex] (we multiplied by [tex]|x-2|\gt 0[/tex])
[tex]x-2\gt \frac{1}{3} \Rightarrow x\gt \frac{7}{3}[/tex]; [tex]x-2\lt -\frac{1}{3} \Rightarrow x\lt \frac{5}{3}[/tex]
Answer: [tex]x\in\left(-\infty, \frac{5}{3}\right)\cup\left(\frac{7}{3}, +\infty\right)[/tex].

Solution:
[tex]\frac{1}{2}|5-x|\lt \frac{3}{2} \Leftrightarrow |x-5|\lt 3 \Leftrightarrow 2\lt x\lt 8[/tex]
The smallest integer in [tex](2, 8)[/tex] is [tex]3[/tex] ([tex]2[/tex] is not included).

Solution:
[tex]|A|\lt B \Leftrightarrow -B\lt A\lt B[/tex] (this also covers the case [tex]B\leq 0[/tex], when there are no solutions):
[tex]x-4\lt 2x+3\lt 4-x[/tex]
[tex]x-4\lt 2x+3 \Leftrightarrow x\gt -7[/tex]; [tex]2x+3\lt 4-x \Leftrightarrow x\lt \frac{1}{3}[/tex]
Answer: [tex]x\in\left(-7, \frac{1}{3}\right)[/tex].

Solution:
[tex]-1\lt x^2-3\lt 1 \Leftrightarrow 2\lt x^2\lt 4[/tex]
[tex]x^2\gt 2 \Leftrightarrow |x|\gt\sqrt{2}[/tex]; [tex]x^2\lt 4 \Leftrightarrow |x|\lt 2[/tex]
Answer: [tex]x\in\left(-2, -\sqrt{2}\right)\cup\left(\sqrt{2}, 2\right)[/tex].
Note: [tex]|A|\lt c[/tex] means a system (both inequalities hold at the same time): [tex]A\gt -c[/tex] and [tex]A\lt c[/tex].

Solution:
Use [tex]x^2=|x|^2[/tex]. Let [tex]t=|x|\geq 0[/tex]:
[tex]t^2-8t+15=(t-3)(t-5)\geq 0 \Leftrightarrow t\leq 3[/tex] or [tex]t\geq 5[/tex]
[tex]|x|\leq 3 \Leftrightarrow -3\leq x\leq 3[/tex]; [tex]|x|\geq 5 \Leftrightarrow x\leq -5[/tex] or [tex]x\geq 5[/tex]
Answer: [tex]x\in(-\infty, -5]\cup[-3, 3]\cup[5, +\infty)[/tex].

Solution:
Check each value:
[tex]x=3:\ |2+9|=11[/tex], [tex]|9+3-4|=8[/tex]; [tex]11\lt 8[/tex] is false.
[tex]x=4:\ 14\lt 16[/tex]; [tex]x=5:\ 17\lt 26[/tex]; [tex]x=6:\ 20\lt 38[/tex] — true.
Answer: [tex]x=3[/tex].