Week 2 — Lecture Outline · Linear Equations & Inequalities

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objectives covered: Objective 2 — Solve linear equations, linear inequalities, and absolute-value equations and inequalities in one variable, and interpret solutions in context.
SLOs touched: A (apply procedures accurately) · B (connect symbolic/numerical representations and interpret in context)
Meeting pattern: 2 sessions × 75 min = 150 min. Segment minutes below total ~150; scale to your own pattern.

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Week at a Glance

The week's big question	"How do we find the value of a variable that makes an equation — or an inequality — true, and how do we communicate what that solution means?"
By the end of the week, students can…	(1) Solve a linear equation in one variable, including clearing fractions, distributing, and collecting variables on one side, and identify identities (infinitely many solutions) and contradictions (no solution); (2) Solve a linear inequality, flip the sign at the right moment, and write the answer in interval notation; (3) Solve an absolute-value equation using the split rule (
Key vocabulary	linear equation, solution, identity, contradiction, properties of equality, isolate, LCD (least common denominator), inequality, solution set, interval notation, open/closed interval, union, compound inequality, absolute value, absolute-value equation, absolute-value inequality
Materials	slides (Deck 2), the week's readings + video links, Desmos (or GeoGebra) and a calculator, one approved chatbot (Gemini / Claude / ChatGPT) for the AI-critique moment and the tutorial
Timing note	8 segments, ~150 min total. Session 1 = Segments 1–4 (~75 min). Session 2 = Segments 5–8 (~75 min).
Holiday note	Labor Day is Monday Sep 7 — no class. Week 2 begins Tuesday Sep 8.

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Segment 1 — Hook & the Promise (7 min) · Session 1 opens

Hook. "Last week you learned the rules that let you rewrite any expression. But here's the thing — knowing the rules isn't the end goal. The goal is to find the number that makes a statement true."

• Write on the board: 3x + 5 = 20. "What is x? How do you find it? And once you can do that — what if the statement isn't equals but greater than? Or involves an absolute value?" Let those questions sit.
• "Every equation and inequality problem you'll ever see in algebra — this term, in calculus, in statistics — traces back to the moves we're going to make this week."

The promise (write it on the board): "By the end of this week, you can solve any linear equation (even one with fractions), read and write interval notation, handle an inequality without forgetting the sign flip, and know exactly what to do when you see absolute value bars."

Why it matters line (memory hook): "Solving is the move. Everything else in algebra sets up a problem so you can make this move."

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Segment 2 — Solving Linear Equations in One Variable (28 min)

Plain language first. A linear equation in one variable is any equation that can be written as ax + b = c (with a ≠ 0). It has exactly one solution — unless it's an identity (always true, infinitely many solutions) or a contradiction (never true, no solution).

The method (one principle, four tactics):
- Principle: whatever you do to one side, you do to the other.
- Tactic 1 — Distribute to clear all parentheses.
- Tactic 2 — Clear fractions by multiplying every term by the LCD.
- Tactic 3 — Collect variables on one side, constants on the other.
- Tactic 4 — Divide both sides by the coefficient of x.

Memory hook: "Distribute, clear fractions, collect, divide — then check."

Fully worked example 1 (distributing, variables on both sides):

Solve 3(x − 2) = 2x + 5.
1. Distribute: 3x − 6 = 2x + 5
2. Subtract 2x from both sides: x − 6 = 5
3. Add 6 to both sides: x = 11
4. Check: 3(11 − 2) = 3(9) = 27 and 2(11) + 5 = 27. ✓
Common error: distributing 3 to (x − 2) as 3x − 2. You must reach every term.

Fully worked example 2 (fraction clearing):

Solve x/2 + 1/3 = 5/6.
1. LCD of 2, 3, 6 is 6. Multiply every term by 6:
6·(x/2) + 6·(1/3) = 6·(5/6) → 3x + 2 = 5
2. Subtract 2: 3x = 3
3. Divide by 3: x = 1
4. Check: 1/2 + 1/3 = 3/6 + 2/6 = 5/6. ✓

Naming the special cases:
- Identity: 2(x + 3) = 2x + 6 → 2x + 6 = 2x + 6 → 0 = 0. True for all x → infinitely many solutions. (This is NOT "one solution" — a common misconception.)
- Contradiction: 2x + 3 = 2x + 7 → 3 = 7. Never true → no solution. Write ∅.

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Segment 3 — Linear Inequalities & Interval Notation (25 min)

Plain language first. An inequality is like an equation, but instead of saying "equals," it says "less than," "greater than," "at most," or "at least." The solution is not one number — it's a range of numbers.

The flip rule (the week's #1 trap — name it explicitly):

Whenever you multiply or divide both sides of an inequality by a negative number, you MUST flip the inequality sign.
Why? 4 > 2, but multiply both sides by −1: −4 < −2. The order reverses.
Write it at the top of your scratch paper: ÷ or × negative → FLIP.**

Interval notation (introduce cleanly):
| Words | Symbol | Interval | Number line |
|---|---|---|---|
| x > a | strict, open at a | (a, ∞) | open circle at a |
| x ≥ a | includes a | [a, ∞) | closed circle at a |
| x < a | strict, open at a | (−∞, a) | open circle at a |
| x ≤ a | includes a | (−∞, a] | closed circle at a |
| a < x ≤ b | between | (a, b] | open left, closed right |

Memory hook: "Parenthesis = peek through the hole (excluded); bracket = blocked, the endpoint is in."

Fully worked example (sign flip):

Solve −2x + 1 < 9. Write the solution in interval notation.
1. Subtract 1: −2x < 8
2. Divide by −2 (negative! → flip < to > ): x > −4
3. Interval notation: (−4, ∞)
Classic wrong answer: x < −4 (flipping the division but not the sign). Say it out loud before every step: "Am I dividing by a negative?"

Matching practice (think-pair-share ~5 min):
Four inequalities on the board; students match to interval notation:
x > −4 → (−4, ∞); x ≤ 3 → (−∞, 3]; −1 < x ≤ 2 → (−1, 2]; x ≥ 0 → [0, ∞). Debrief: which endpoint gets a bracket and why?

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Segment 4 — Misconceptions + Quick Interaction (15 min) · Session 1 closes (~75 min)

Name the misconceptions out loud, then cure each:

• ❌ "−3x > 9 → x > −3." (Divided by −3 without flipping.)
  ✅ Cure: −3x > 9 → x < −3. Dividing by −3 (negative) flips the sign. Check: try x = −4: −3(−4) = 12 > 9 ✓. Try x = 0: −3(0) = 0 ≯ 9 ✗. The flip is correct.

• ❌ "2(x + 3) = 2x + 6 has exactly one solution."
  ✅ Cure: Simplify to 0 = 0 — this is an identity, true for every real number. The answer is "infinitely many solutions," not "one solution."

• ❌ "A square bracket goes on −∞."
  ✅ Cure: infinity is not a number; it's a direction. We can never "reach" it, so it always gets a parenthesis: (−∞, ...) and (..., ∞).

• ❌ "Sign errors when distributing a negative into an inequality."
  ✅ Cure: work on equations and inequalities the same way — distribute first, then handle the inequality. The flip rule only applies at the divide/multiply step.

Think-Pair-Share — 2 min each:
- "Your classmate solved −5x > 15 and got x > −3. Where did they go wrong and what is the right answer?" (Should be x < −3.)
- "Is 3x − 4 = 3(x − 4) + 8 an identity, a contradiction, or does it have one solution?" (Simplify: 3x − 4 = 3x − 12 + 8 = 3x − 4 → identity.)

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Segment 5 — Absolute-Value Equations (20 min) · Session 2 opens

Hook back in: "Last session we solved inequalities with regular expressions. Now we'll meet a twist — what happens when the variable is inside absolute value bars?"

Plain language first. The absolute value |X| = distance from 0 on the number line — always non-negative. |3| = 3 and |−3| = 3.

The split rule for |X| = a:
- If a < 0: no solution (distance can't be negative).
- If a = 0: X = 0 (one solution).
- If a > 0: two solutions — X = a or X = −a.

Memory hook: "Absolute value = distance. Negative distance is impossible. Zero distance means you're at zero. Positive distance means two spots — one on each side."

Fully worked example 1:

Solve |x| = 7.
a = 7 > 0 → x = 7 or x = −7. Solution: {−7, 7}.

Fully worked example 2 (variable expression inside):

Solve |2x − 1| = 7.
a = 7 > 0 → two cases:
- Case 1: 2x − 1 = 7 → 2x = 8 → x = 4
- Case 2: 2x − 1 = −7 → 2x = −6 → x = −3
Check: |2(4) − 1| = |7| = 7 ✓; |2(−3) − 1| = |−7| = 7 ✓.

Name the misconception:
- ❌ "|x| = −5 has the solution x = ±5."
✅ Cure: the right side is −5 < 0. A distance cannot be negative. No solution. The common error is mechanically writing ±(−5) = ±5 without checking the sign of a first.

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Segment 6 — Absolute-Value Inequalities (22 min)

Plain language first. The compound-inequality rules come directly from the meaning of distance.

Rule 1 — "Less than" (AND — a bounded interval):

|X| < a (with a > 0) → −a < X < a.
"X is within a of zero" — both directions, so it's a closed interval between −a and a.
Memory hook: |X| < a → "middle" — think AND, the solution is BETWEEN the endpoints.

Rule 2 — "Greater than or equal" (OR — two rays):

|X| ≥ a (with a > 0) → X ≤ −a or X ≥ a.
"X is at least a away from zero" — two separate tails.
Memory hook: |X| > a → "outside" — think OR, the solution is on BOTH SIDES going outward.

The "and"/"or" confusion (name it explicitly — biggest AV error):

❌ "|x| > 3 → −3 < x < 3" (treating > as AND).
✅ |x| > 3 → x < −3 or x > 3. The solution is the two outer rays, not the middle.

Fully worked example 1 (less-than → AND):

Solve |x| < 5. Write in interval notation.
a = 5 > 0, "less than" → middle: −5 < x < 5 → (−5, 5).

Fully worked example 2 (greater-than-or-equal → OR):

Solve |x| ≥ 3. Write in interval notation.
a = 3 > 0, "greater than or equal" → outside: x ≤ −3 or x ≥ 3 → (−∞, −3] ∪ [3, ∞).

Quick check (show of hands or clicker):
"If |2x + 1| < 7, does the solution graph look like a single segment or two rays?"
→ single segment (less-than → AND → bounded). Let that picture guide the rule.

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Segment 7 — Technology Workflow + AI-Critique Moment (15 min)

Technology workflow — graph a linear inequality in Desmos:
1. Open desmos.com/calculator.
2. Type y = 2x + 1 — this is the line x = (y − 1)/2. Better: type the original equation as an inequality: in Desmos type y > 2x + 1 — it shades the region where the inequality holds. (For one-variable inequalities, use the number line view.)
3. For checking an absolute-value inequality like |2x − 1| < 7: type y = |2x-1| on line 1, and y = 7 on line 2. The x-values where the V-curve is below the horizontal line are your solution.

AI-critique moment:

Paste to an approved chatbot: "Solve −3x > 9 and write the answer in interval notation."
The correct answer is x < −3, i.e. (−∞, −3).
Chatbots frequently forget to flip the inequality sign and return x > −3, i.e. (−3, ∞). Check the model's work by hand and correct it if wrong. This is the Week 2 trap in the wild.

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Segment 8 — Signature Trap Slide, Callback & Hand-off (13 min) · Session 2 closes (~75 min)

The Signature Trap (give it dedicated time):

The forgotten flip.
A student solves −3x > 9:
Step 1: Divide both sides by −3 → x > −3. ← WRONG
What went wrong: dividing by a negative requires flipping the sign.
Step 1 (corrected): Divide both sides by −3 → x < −3. ← RIGHT
"Every time you see a negative coefficient to divide away, say 'flip' out loud before you write the sign."

Callback + tease:
- Callback: "Every step we took this week — isolating a variable, distributing, tracking signs — is Week 1's rules in action. The properties from Week 1 are why these moves are legal."
- Tease next week: "Next week we zoom out: instead of one equation with one solution, we'll look at relationships between two variables and the whole picture of a function."

Hand-off (the week's graded work):
- Lecture Tutorial 2 (AI tutor, share-link submission) — linear equations, inequalities, interval notation, absolute-value equations and inequalities.
- Quiz 2 (end of week, no AI) and Discussion 2 ("Spot the Sign Error" — error-analysis on the forgotten flip).
- Assignment 2 ("Solving Is the Move") — AI-coached, self-scored.

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Instructor FAQ — Common Stumbles

Student says / does	Quick cure
Divides by −3 but doesn't flip the sign.	Say it every single time: "negative divisor → flip." Put it on a sticky note. Check with a test value after solving.
Writes x < −4 when the answer is x > −4 (or vice versa).	Have them test a value in the original. If x = 0 works, the solution should include 0.
Writes "2(x+3)=2x+6 has one solution."	Simplify all the way: if you reach 0 = 0, it's an identity (infinitely many). "One solution" means you get x = some specific number.
Thinks	x
Uses "and" for	x
Uses "or" for	x
Writes ∞ with a bracket: [3, ∞].	Infinity is not a number — we never "reach" it. Always use a parenthesis: [3, ∞).
Leaves the answer as an inequality instead of interval notation.	The two notations must be fluent: x > −4 and (−4, ∞) say the same thing. Practice writing both until either one flows automatically.

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Scope flag

This outline stays within Objective 2. Compound inequalities joined by AND/OR beyond the absolute-value context (e.g., 1 < x + 2 < 7) are touched through the absolute-value connection but not drilled separately — they are not in the scope of this objective. Applications are kept to single-step translations (taxi fare, rental cost) to reinforce the concept without competing for time. The identity/contradiction classification is intentional depth — it appears on the Quiz 2 True/False item and recurs in every equation-solving context through the term.