Compound and Absolute Value Inequalities

⛓️ Compound & Absolute Value Inequalities

Part 1 of 5 — Compound Inequalities: AND vs. OR

Topics in This Part

🔑 Key Concept: A compound inequality joins two simple inequalities with the word and or the word or. The joining word changes everything — which numbers count as solutions, and what the graph looks like.

What Is a Compound Inequality?

A single inequality like $x > 3$ describes one condition. A compound inequality describes two conditions at once:

• AND — a number must satisfy both parts. (Think: the overlap.)
• OR — a number must satisfy at least one part. (Think: either side counts.)

The Two Flavors

💡 The "between" shortcut: An AND inequality where $x$ is squeezed between two numbers can be written in one line. "$x > -1$ and $x < 4$" becomes $-1 < x < 4$.

Concept Check 🎯

Reading the Number Line

The graph instantly tells you AND from OR:

• AND graphs as a single connected segment — the overlap of the two pieces. (Often a finite chunk in the middle.)
• OR graphs as two rays pointing away from each other — the union of the pieces.

Open vs. Closed Dots

Example — AND: $-1 < x \le 4$ graphs as an open dot at $-1$, a closed dot at $4$, and the segment between them shaded.

Example — OR: $x < -2 \;\text{or}\; x \ge 5$ graphs as a ray going left from an open dot at $-2$, plus a ray going right from a closed dot at $5$.

AND or OR? 🔽

Decide what kind of solution set each compound inequality produces.

Writing a Compound Inequality

Going the other way — from a number line to symbols — comes up constantly.

• An AND segment from $-2$ (closed) to $5$ (open) is written $-2 \le x < 5$. Read it left-to-right: the left number ties to the left end.
• An OR pair (ray left from $1$, ray right from $4$) is written $x < 1 \;\text{or}\; x > 4$. Each ray gets its own piece.

💡 In a three-part AND inequality, the smaller number is always on the left. If yours reads "$5 < x < -2$", you've reversed it.

Name the Bounds 🧮

A graph shows a shaded segment with a closed dot at $-4$ and an open dot at $3$ (everything between shaded). It represents $a \le x < b$.

1) Enter $a$ (left bound). 2) Enter $b$ (right bound).

Putting It Together

You now know the single most important fork in this whole topic:

🔑 AND = overlap (intersection), OR = combine (union). Keep this front and center — every problem in Parts 2–5 comes back to it.

In Part 2 we'll actually solve compound inequalities — moving terms around while keeping the AND/OR logic intact.

⛓️ Compound & Absolute Value Inequalities

Part 2 of 5 — Solving Compound Inequalities

🔑 The Golden Rule: Solve an inequality exactly like an equation — except when you multiply or divide by a negative number, flip the inequality sign.

Solving a "Between" (AND) Inequality

When $x$ is sandwiched in the middle, do the same operation to all three parts at once.

Worked Example: $-7 < 2x + 1 \le 9$

1. Subtract $1$ from all three parts: $-8 < 2x \le 8$
2. Divide all three parts by $2$ (positive, so no flip): $-4 < x \le 4$

So the solution is $-4 < x \le 4$: an open dot at $-4$, closed dot at $4$, shaded between.

✅ Check $x = 0$: $2(0)+1 = 1$, and $-7 < 1 \le 9$ ✓.

The Sign-Flip Trap ⚠️

Worked Example: $-4 \le -2x < 6$

Divide all three parts by $-2$. Dividing by a negative flips both inequality signs:

$\frac{-4}{-2} \;\ge\; x \;>\; \frac{6}{-2} \quad\Longrightarrow\quad 2 \ge x > -3$

Rewritten smallest-to-largest: $-3 < x \le 2$.

⚠️ Most common mistake: forgetting to flip — and forgetting that when you flip, the whole chain reverses direction. Always re-read your final answer left-to-right to make sure it still makes sense.

Concept Check 🎯

Solving an "OR" Inequality

An OR inequality is really two separate inequalities. Solve each one on its own, then combine.

Worked Example: $3x - 1 < -7 \;\text{or}\; 2x + 5 > 11$

Solve each piece: $3x - 1 < -7 \;\Rightarrow\; 3x < -6 \;\Rightarrow\; x < -2$ $2x + 5 > 11 \;\Rightarrow\; 2x > 6 \;\Rightarrow\; x > 3$

Final answer: $x < -2 \;\text{or}\; x > 3$ — two rays pointing apart.

💡 You cannot write an OR result as a single three-part inequality. Three-part notation is reserved for AND ("between") solutions only.

Track the Steps 🔽

You're solving $-7 < 2x + 1 \le 9$. Choose the result of each move.

Always Check One Point

After solving, plug a number from inside your answer back into the original inequality. If it works, your solution set is on the right track.

For $-4 < x \le 4$, try $x = 0$ in $-7 < 2x + 1 \le 9$: $\;2(0)+1 = 1$, and $-7 < 1 \le 9$ ✓.

💡 This one habit catches almost every sign-flip and arithmetic slip. Now try the drills yourself.

Solve It 🧮

1) Solve $-3 < x - 2 < 5$. Enter the lower bound, then the upper bound of $x$. 2) Solve $4x \ge 12$ (the right piece of an OR). Enter the bound: $x \ge \,?$

⛓️ Compound & Absolute Value Inequalities

Part 3 of 5 — Absolute Value as Distance

🔑 The Big Idea: $|x|$ means the distance from $x$ to $0$ on the number line. Distance is never negative — and that single fact powers every absolute value inequality.

Absolute Value = Distance

$|x| = 5$ asks: which numbers are exactly $5$ units from $0$? Answer: $5$ and $-5$.

Inequalities ask about distance being small or large:

🔑 "LESS than" → AND (a squeezed-in segment). "GREATER than" → OR (two rays). A handy memory hook: "Less thAND, greatOR."

Concept Check 🎯

Two Special Cases ⚠️

Because absolute value is never negative, watch for these:

⚠️ Don't auto-pilot the split! If the right side is negative, stop and think about distance first. $|x| < (\text{negative})$ is impossible; $|x| > (\text{negative})$ is always true.

Translate the Statement 🔽

Rewrite each absolute value inequality. Watch for the special cases.

Keeping the Endpoint Symbol

When you rewrite an absolute value inequality, the relation symbol carries over to both new pieces:

💡 A closed symbol ($\le, \ge$) stays closed; an open symbol ($<, >$) stays open. The endpoints just become $-c$ and $c$.

Find the Boundaries 🧮

Rewrite $|x| \le 9$ as a compound inequality $a \le x \le b$.

1) Enter $a$ (the lower bound). 2) Enter $b$ (the upper bound).

⛓️ Compound & Absolute Value Inequalities

Part 4 of 5 — Solving |ax + b| Inequalities

🔑 The Method: Split the absolute value into a compound inequality, then solve like Part 2. "Less thAND, greatOR" tells you which kind you get.

The "Less Than" Case → AND

To solve $|ax + b| < c$ (with $c > 0$), rewrite it as the single AND chain:

$-c < ax + b < c$

then solve all three parts at once.

Worked Example: $|2x - 3| \le 7$

Split: $-7 \le 2x - 3 \le 7$.

1. Add $3$ to all parts: $\;-4 \le 2x \le 10$
2. Divide by $2$: $\;-2 \le x \le 5$

Solution: $-2 \le x \le 5$ (a closed segment).

✅ Check $x = 5$: $|2(5)-3| = |7| = 7 \le 7$ ✓.

The "Greater Than" Case → OR

To solve $|ax + b| > c$ (with $c > 0$), rewrite it as two inequalities joined by OR:

$ax + b < -c \quad\text{or}\quad ax + b > c$

Worked Example: $|3x + 1| > 8$

Split into two: $3x + 1 < -8 \;\Rightarrow\; 3x < -9 \;\Rightarrow\; x < -3$ $3x + 1 > 8 \;\Rightarrow\; 3x > 7 \;\Rightarrow\; x > \tfrac{7}{3}$

Solution: $x < -3 \;\text{or}\; x > \tfrac{7}{3}$ (two rays).

⚠️ For the OR case, notice the left piece keeps the negative of $c$ and flips to $<$. Don't just copy "$> 8$" twice.

Set Up the Split 🔽

Choose the correct first step (the compound form) for each.

The Three-Step Routine

Every $|ax + b|$ inequality follows the same routine:

1. Decide AND or OR using "Less thAND, greatOR."
2. Split into the matching compound form.
3. Solve the resulting inequality (Part 2 skills — and flip on a negative divide).

⚠️ One more reflex: if the right-hand side is negative, skip the split and use the special cases from Part 3 ($<$ negative → no solution; $>$ negative → all reals).

Concept Check 🎯

Picture the Answer

A quick sketch catches sign errors:

• AND answers (from $<$) shade a single segment around the center. The center is the value that makes the inside zero — e.g. for $|x-1|$ that's $x = 1$.
• OR answers (from $>$) shade two rays pointing away from that center.

💡 If your "less than" problem gives two rays, or your "greater than" gives a segment, you've mixed up AND and OR — go back to "Less thAND, greatOR."

Solve It 🧮

1) Solve $|x - 1| \le 4$, written as $a \le x \le b$. Enter $a$, then $b$. 2) Solve $|x + 3| > 5$ → the right ray is $x > \,?$. Enter that bound.

⛓️ Compound & Absolute Value Inequalities

Part 5 of 5 — Applications & Mastery Check

You can now (1) tell AND from OR, (2) solve compound inequalities, (3) read absolute value as distance, and (4) solve $|ax+b|$ inequalities. Let's apply it and finish strong.

Real-World: Tolerance Problems

Manufacturing and science use absolute value to describe how far a measurement may stray from a target.

Example: A bolt should be $50$ mm long, with a tolerance of $0.4$ mm.

"Within $0.4$ of the target" is a distance, so we write:

$|L - 50| \le 0.4$

Split (less than → AND): $\;-0.4 \le L - 50 \le 0.4$, then add $50$:

$49.6 \le L \le 50.4$

So an acceptable bolt is between $49.6$ mm and $50.4$ mm long.

💡 Pattern to memorize: $|{\text{actual}} - {\text{target}}| \le {\text{tolerance}}$. The target is what you subtract; the tolerance is the right-hand side.

Tolerance Range 🧮

A cereal box should hold $500$ g, with a tolerance of $6$ g. Acceptable weights $W$ satisfy $|W - 500| \le 6$, which gives $a \le W \le b$.

1) Enter $a$ (the minimum acceptable weight, in grams). 2) Enter $b$ (the maximum acceptable weight, in grams).

Quick Reference

⚠️ Two reflexes to keep: flip the sign when you multiply/divide by a negative, and check the special cases before splitting when the right side is negative.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.