Exercises — Compound inequalities — AND, OR

L1.1 The word is and → ==intersection $\cap$==. We want numbers bigger than 2 and smaller than 9 — the overlap. Both endpoints are strict ($>$, $<$), so both are open circles → $(2,9)$.

L1.2 The word is or → ==union $\cup$==. Keep everything either ray covers. These rays point apart and don't overlap, so we get two pieces: $(-\infty ,-1]\cup [4,\infty )$. Both endpoints are $\le$ / $\ge$, so they are included (square brackets).

L1.3 False. The compact three-part form $a<x<b$ always means AND (both $3<x$ and $x<8$ hold at once). There is no compact form for OR — you must write two separate inequalities.

L2.1 Solve each piece (each is a Linear inequalities in one variable):

• $x+2>5\Rightarrow x>3$, so $A=(3,\infty )$.
• $x-1\le 6\Rightarrow x\le 7$, so $B=(-\infty ,7]$.

AND → intersection → numbers $>3$ and $\le 7$ → $\boxed{(3,7]}$. Compact: $3<x\le 7$. (Round on $3$ = excluded, square on $7$ = included.)

L2.2 Solve each piece:

• $2x<-4\Rightarrow x<-2$, so $A=(-\infty ,-2)$.
• $x-3\ge 2\Rightarrow x\ge 5$, so $B=[5,\infty )$.

OR → union → keep both pieces. They don't overlap → $\boxed{(-\infty ,-2)\cup [5,\infty )}$.

L2.3 Same operation to all three parts at once.

• Add $3$ everywhere: $1+3\le 2x<9+3\Rightarrow 4\le 2x<12$.
• Divide everywhere by $2$ (positive — no flip): $2\le x<6$.

Answer: $\boxed{[2,6)}$.

L3.1 $A=(6,\infty )$ points right; $B=(-\infty ,2)$ points left. They aim away from each other, so there is no number that is both $>6$ and $<2$. AND → intersection → $\boxed{\varnothing }$ (empty set, no solution).

L3.2 $A=(-\infty ,7)$ and $B=(1,\infty )$ overlap on $(1,7)$ and together spill over the whole line. Pick any real number $r$: it is either less than $7$ or greater than $1$ (or both). So OR → union → $\boxed{\mathbb{R}=(-\infty ,\infty )}$.

L3.3 This is the degenerate single-point case. $A=(-\infty ,4]$ reaches up to and includes $4$; $B=[4,\infty )$ starts at and includes $4$. Their overlap is only the one number they share: $(-\infty ,4]\cap [4,\infty )=\{4\}.$ Because both endpoints are included ($\le$ and $\ge$), the single point $4$ survives. In interval notation this shrivelled interval is written $\boxed{[4,4]}$, which is just the set $\{4\}$. (Had either inequality been strict — say $x<4$ AND $x\ge 4$ — the point $4$ would be excluded by the strict side, and the overlap would collapse to $\varnothing$.)

L3.4 Solve all three parts together, then handle the negative divisor carefully.

• Start: $-4<-2x\le 6$.
• Divide every part by $-2$. Dividing by a negative flips BOTH relation signs at once:

$\frac{-4}{-2}>\frac{-2x}{-2}\ge \frac{6}{-2}\Rightarrow 2>x\ge -3.$

• Why rewrite the order? The chain $2>x\ge -3$ reads "right-to-left" (big number on the left), which is correct but awkward — a reader expects the smaller bound on the left. Reversing the entire chain (swap both ends and flip both relation signs together) gives the identical statement written low-to-high: $-3\le x<2$. Nothing changed mathematically; we just made it readable.

Answer: $\boxed{[-3,2)}$.

L4.1 By the Absolute value inequalities rule, $|E|<a$ (with $a>0$) means the distance of $E$ from $0$ is less than $a$, i.e. $E$ is trapped between $-a$ and $a$. That "trapped between" is an AND: $|x-3|<5⟺-5<x-3<5.$ Add $3$ to all three parts: $-2<x<8$. Answer: $\boxed{(-2,8)}$. Connective: AND.

L4.2 For $|E|\ge a$, the distance is at least $a$, so $E$ sits outside the band — either far left or far right. That "either/or" is an OR: $|2x+1|\ge 7⟺2x+1\le -7\text{ OR }2x+1\ge 7.$

• Left piece: $2x+1\le -7\Rightarrow 2x\le -8\Rightarrow x\le -4$.
• Right piece: $2x+1\ge 7\Rightarrow 2x\ge 6\Rightarrow x\ge 3$.

Answer: $\boxed{(-\infty ,-4]\cup [3,\infty )}$. Connective: OR.

L4.3 Same operation to all three parts.

• Multiply every part by $2$: $-6\le 4-x<2$.
• Subtract $4$ everywhere: $-10\le -x<-2$.
• Multiply by $-1$ (flip both signs): $10\ge x>2$.
• Why rewrite the order? $10\ge x>2$ is correct but written high-to-low, which is hard to read. Reverse the whole chain at once — swap both ends and flip both relation signs together — to get the same statement low-to-high: $2<x\le 10$. This is purely cosmetic; the solution set is unchanged.

Answer: $\boxed{(2,10]}$.

L5.1 "Taller than 120" → $h>120$. "No taller than 195" → $h\le 195$. Both must hold → AND: $120<h\le 195.$ Answer: $\boxed{(120,195]}$ cm. (Round on $120$ = excluded, square on $195$ = included.)

L5.2 Two steps.

• "Charges fine" set: $|T-22|\le 18⟺-18\le T-22\le 18⟺4\le T\le 40$, i.e. $[4,40]$.
• "Does NOT fast-charge" set: we want the phone to refuse fast charging, which happens when $T<10$ or $T>35$. So this set is $(-\infty ,10)\cup (35,\infty )$.
• Intersect the two (charges AND not fast-charging):
  • $[4,40]\cap (-\infty ,10)=[4,10)$ — the low band.
  • $[4,40]\cap (35,\infty )=(35,40]$ — the high band.

Answer: $\boxed{[4,10)\cup (35,40]}$ °C. (See Set operations — union and intersection for the intersect-then-union bookkeeping.)

L5.3 Use OR to remove a single point: $x<1\text{ OR }x>1.$ The union of $(-\infty ,1)$ and $(1,\infty )$ is every real number except $1$. ✔

Why AND cannot do this: an AND takes the intersection of two sets, which is always one contiguous overlap (or empty, or a single point). It can only ever produce a single interval (or $\varnothing$) — it can never leave a hole in the middle of the line. Punching out an interior point requires combining two separated rays, and only union (OR) does that.