Week 5 — Lecture Outline · Systems of Linear Equations & Inequalities

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objectives covered: Objective 4 — Graph and analyze linear functions and solve systems of linear equations and inequalities.
SLOs touched: A (apply procedures accurately) · B (connect symbolic/graphical 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	"When two linear equations describe the same world, how do we find the point — if there is one — where both are true at once?"
By the end of the week, students can…	(1) solve a 2×2 linear system by substitution; (2) solve a 2×2 linear system by elimination (including when a multiply is needed); (3) classify a system as consistent independent, inconsistent, or dependent and state the geometric meaning; (4) graph and interpret a system of linear inequalities — shading both half-planes, identifying the overlap region, and testing a point.
Key vocabulary	system of equations, solution of a system, substitution, elimination, consistent, inconsistent, dependent, independent, half-plane, boundary line, solution region, test point
Materials	slides (Deck 5), week's readings + videos, Desmos (or GeoGebra) for graphing and checking, 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 (~73 min). Session 2 = Segments 5–8 (~77 min).

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Segment 1 — Hook & the Promise (8 min) · Session 1 opens

Hook. "Raise your hand if you've ever split a cost with someone — maybe rent, maybe a shared dinner — and figured out what each person owed based on two pieces of information: what the total was, and how much more one person used than the other."

Wait for hands — almost every hand goes up.

• "You just described a system of two equations. You had two unknowns — the two amounts — and two rules about them. The algebra we build this week is the precise machinery for solving that problem every time, for any two-equation, two-unknown situation."
• "This week is also where algebra connects visibly to geometry. Two linear equations are two lines on the same graph. Their solution — if it exists — is the point where they cross."

The promise (write it on the board): "By the end of this week you can take any system of two linear equations — or two linear inequalities — and solve it, classify it, and interpret it. And you'll never again confuse 'no solution' with 'infinitely many.' They look almost the same in the algebra — but they're geometrically completely different."

Why it matters (memory hook): "Every optimization problem in business, every equilibrium in economics, and every break-even point in accounting is a system of equations. Systems are where linear algebra meets the real world."

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Segment 2 — Solving by Substitution (22 min)

Plain language first. Substitution means: isolate one variable in one equation, then substitute that expression into the other equation — replacing one unknown with something you already know. Now you have one equation in one unknown, which you can solve.

The three-step recipe:
1. Isolate — choose the equation where one variable is easiest to isolate. Solve for that variable.
2. Substitute — replace that variable in the other equation with the expression you found.
3. Back-substitute — once you have one value, plug it into either original equation to find the second value. Check in both.

Memory hook: Substitution = replace one unknown with what you know.

One fully worked example (every step):

Solve the system by substitution: y = 2x and x + y = 9.

1. The first equation already isolates y: y = 2x. (Step 1 is done.)
2. Substitute into the second equation: x + (2x) = 9 → 3x = 9 → x = 3.
3. Back-substitute: y = 2(3) = 6.
4. Check: does (3, 6) satisfy both equations?
   - y = 2x: 6 = 2(3) = 6 ✓
   - x + y = 9: 3 + 6 = 9 ✓
5. Solution: (3, 6).

What it means geometrically: the lines y = 2x and x + y = 9 intersect at the point (3, 6).

Common errors to name and cure:
- ❌ Substituting into the same equation you solved, not the other. → ✅ Always substitute into the equation you did NOT isolate from.
- ❌ Forgetting to back-substitute. → ✅ You have x = 3, but the solution is an ordered pair (x, y); you must find y too.
- ❌ Sign error when substituting a negative expression. → ✅ If y = −2x + 5 and you substitute, write x + (−2x + 5) — use parentheses every time.

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Segment 3 — Solving by Elimination (25 min)

Plain language first. Elimination means: add the two equations so that one variable literally disappears — is eliminated. For this to work, the coefficients of that variable must be opposites (like +y and −y). If they're not, multiply one or both equations by a number to make them opposites first.

The four-step recipe:
1. Align — write both equations in standard form (Ax + By = C) so like terms line up.
2. Multiply (if needed) — multiply one or both equations by a constant so one variable's coefficients are opposites.
3. Add — add the equations; the targeted variable cancels.
4. Solve, then back-substitute — solve for the remaining variable, then plug back to find the other.

Memory hook: Elimination = make a pair of coefficients into opposites, then add.

Example 1 — no multiply needed:

Solve: x + y = 10 and x − y = 4.

1. Aligned already. Note: +y in equation 1, −y in equation 2 — they are opposites.
2. No multiply needed.
3. Add: (x + y) + (x − y) = 10 + 4 → 2x = 14 → x = 7.
4. Back-substitute: 7 + y = 10 → y = 3. Solution: (7, 3).
   Check: 7 + 3 = 10 ✓; 7 − 3 = 4 ✓

Example 2 — multiply first:

Solve: 2x + 3y = 12 and x − y = 1.

1. Aligned: both in Ax + By = C form.
2. Multiply: to eliminate y, multiply equation 2 by 3 → 3x − 3y = 3. Now the y-terms are +3y and −3y.
3. Add: (2x + 3y) + (3x − 3y) = 12 + 3 → 5x = 15 → x = 3.
4. Back-substitute: 3 − y = 1 → y = 2. Solution: (3, 2).
   Check: 2(3) + 3(2) = 6 + 6 = 12 ✓; 3 − 2 = 1 ✓

Common errors:
- ❌ Multiplying only one term, not every term. If you multiply equation 2 by 3, the 1 on the right also becomes 3.
- ❌ Sign error when adding: thinking (+3y) + (+3y) = 0. They only cancel when one is positive and one is negative.

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Segment 4 — Classifying Systems + Misconceptions (18 min) · Session 1 closes (~73)

Plain language first — three possible outcomes when you solve a 2×2 system:

Outcome	Algebra signal	Geometric picture	Classification
One solution	You get a specific (x, y)	Two lines intersect at one point	Consistent, independent
No solution	The variable cancels AND you get a false equation (e.g., 0 = 7)	Two lines are parallel — same slope, different intercepts	Inconsistent
Infinitely many	The variable cancels AND you get a true equation (e.g., 0 = 0)	The two equations describe the same line	Consistent, dependent

Memory hook: "Inconsistent = impossible (parallel lines never meet). Dependent = the same line described twice."

Worked example — inconsistent:

Classify: x + y = 3 and x + y = 7.
Add equations designed to eliminate x: subtract eq 1 from eq 2:
(x + y) − (x + y) = 7 − 3 → 0 = 4. False. → Inconsistent; no solution. (Same slope = 1, different y-intercepts 3 and 7 → parallel lines.)

Worked example — dependent:

Classify: 2x + y = 5 and 4x + 2y = 10.
Notice that equation 2 = 2 × equation 1. Multiply equation 1 by −2 and add:
−2(2x + y = 5) → −4x − 2y = −10. Add: (4x + 2y) + (−4x − 2y) = 10 + (−10) → 0 = 0. True. → Dependent; infinitely many solutions. (Same line.)

Named misconceptions (say each out loud, then cure):
- ❌ "0 = 4 means x = 4." → ✅ 0 = 4 is the signal for no solution — no value of x can make 0 equal 4.
- ❌ "0 = 0 means no solution — zero solutions." → ✅ 0 = 0 is a true statement; it means the system has infinitely many solutions because every point on the line works.
- ❌ "Parallel lines have the same equation." → ✅ Parallel lines have the same slope but different y-intercepts; they share no points. The same line (same slope and same intercept) means the equations are multiples of each other.

Think-Pair-Share (rapid, ~5 min): display three systems; students classify solo (30 sec), share with a neighbor, class votes. Use one of each type.

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Segment 5 — Systems of Linear Inequalities (20 min) · Session 2 opens

Hook back in: "We've been solving systems for equality — both equations hold exactly. What if instead we have inequalities? y > x says y must be more than x, not equal. Two inequalities together define a region of the plane, not just a point."

Plain language first. Each linear inequality has a boundary line (the equation you get by replacing the inequality sign with =). On one side of that line, the inequality is true; on the other, it's false. That true side is the half-plane.

For a system of two inequalities:
1. Graph each boundary line (dashed if strict >, <; solid if ≥, ≤).
2. Shade the half-plane for each inequality (test a point — usually (0, 0) is easiest if it's not on the line).
3. The solution is the overlapping (intersection) region — where both shadings cover.
4. Test a point inside the overlap to confirm it satisfies both inequalities.

Memory hook: "The solution region is where both shadings overlap — the AND of two conditions."

Fully worked example:

Solve and graph the system: y > x and y < 4.

• y > x: boundary is y = x (dashed, since strict). Test (0, 1): 1 > 0 ✓ → shade above y = x.
• y < 4: boundary is y = 4 (dashed). Test (0, 0): 0 < 4 ✓ → shade below y = 4.
• Overlap region: above y = x AND below y = 4.
• Test point (1, 3): y > x → 3 > 1 ✓; y < 4 → 3 < 4 ✓. Confirms (1, 3) is in the solution region.

Other test points to discuss: (2, 2): 2 > 2? No — on the boundary, not in. (3, 5): 5 < 4? No — outside. (0, 0): 0 > 0? No — on the boundary.

Common errors:
- ❌ Shading the wrong side. → ✅ Always test a point. If (0, 0) gives a true statement, shade toward (0, 0).
- ❌ Making the boundary line solid when the inequality is strict (<, >). → ✅ Strict inequality = dashed (the boundary is NOT included).

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Segment 6 — The Signature Trap: Confusing "No Solution" and "Infinitely Many" (15 min)

Name it explicitly. This week's most punishing error: students see that the variable eliminated and think "something went wrong" — but what happened after the variable cancels is the entire story.

The Trap Slide (put exactly this on a board or slide):

What you get after eliminating	What it means	Example
0 = k where k ≠ 0 (false)	No solution (inconsistent)	0 = 4
0 = 0 (true)	Infinitely many solutions (dependent)	0 = 0
x = k (a number)	One solution (consistent, independent)	x = 3

Two common distractors on tests:
1. The elimination sign error: multiplying an equation by a negative and forgetting to flip the right-hand side. Example: −1(2x + y = 5) must give −2x − y = −5, not −2x − y = 5.
2. Confusing parallel slope test: x + y = 3 and x + y = 7 have the same left side and different right sides — two lines with slope −1 and different intercepts. Students sometimes think they must intersect "somewhere far out."

Callback + cure: "Run the algebra — don't guess from the look of the equations. If 0 = k (with k ≠ 0), it's impossible; if 0 = 0, it's the same line."

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Segment 7 — Technology Workflow + AI-Critique Moment (14 min)

Technology workflow — check a system in Desmos (exact steps):
1. Open desmos.com/calculator (free, no login).
2. Type the first equation on line 1, e.g., y = 2x.
3. Type the second equation on line 2, e.g., x + y = 9.
4. The two lines appear. Click their intersection point — Desmos labels it (3, 6). This confirms your algebra.
5. For inequalities: type y > x on line 1 and y < 4 on line 2. Desmos shades each half-plane; the overlap region appears immediately.

AI-critique moment (students verify, not consume):

Paste this to an approved chatbot: "Solve the system by elimination: 3x + 2y = 10 and 3x − 2y = 2. Show every step."

The correct answer: add the equations → 6x = 12 → x = 2; back-substitute → 3(2) + 2y = 10 → 2y = 4 → y = 2; solution: (2, 2).

Common chatbot errors: (1) adding when it should subtract to eliminate, getting 4y = 8 instead; (2) back-substituting into the wrong equation and getting a different y; (3) not checking the solution in both equations. Your job: check the chatbot's steps against your own algebra. The tool drafts; you judge.

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Segment 8 — Callback, Hand-off & Next-Week Tease (8 min) · Session 2 closes (~77)

Callback:
- To Week 4: "Two weeks ago you graphed a single line by slope and intercept. This week you put two lines on the same plane and found where they meet — or confirmed they can't."
- This week's big idea: "Substitution = replace. Elimination = cancel. Both give the same answer; the skill is choosing which is faster for a given system."
- Classification check: "The algebra tells you the geometry: one number → one intersection. A false equation → parallel lines. A true-but-trivial equation → same line."

Hand-off (this week's graded work):
- Lecture Tutorial 5 — substitution, elimination, classifying, systems of inequalities.
- Quiz 5 (end of week, no AI) and Discussion 5 ("Substitution or Elimination?" — method choice and reasoning).
- Assignment 5 ("Two Equations, One Answer") — AI-coached, self-scored.

Tease next week:
- "Week 6 we leave the linear world. Instead of first-degree terms, we'll have x², and instead of straight lines, we'll have parabolas. Polynomials and factoring — the machinery that powers quadratics — are next."

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Instructor FAQ — Common Stumbles

Student says / does	Quick cure
Substitutes into the equation they just solved (not the other one).	After isolating y in equation 1, substitute into equation 2. Substituting back into equation 1 just reproduces the same equation and gives 0 = 0 whether or not that's true.
Gets x = 3 and calls (3, ?) the solution — never finds y.	The solution to a system is an ordered pair (x, y). After finding x, always back-substitute to find y.
Sees 0 = 0 and says "no solution."	0 = 0 is always true → infinitely many solutions (dependent). "No solution" comes from 0 = k with k ≠ 0.
Forgets to multiply the right-hand side when scaling an equation.	Multiplying "both sides" means the constant on the right gets multiplied too. −2(2x + y = 5) → −4x − 2y = −10, not −4x − 2y = 5.
Shades the wrong half-plane.	Test (0, 0). If (0, 0) satisfies the inequality, shade toward (0, 0); otherwise shade away. The test point never lies — assuming (0, 0) isn't on the boundary.
Draws a solid line for a strict inequality.	Strict < or > → dashed line (boundary not included). Non-strict ≤ or ≥ → solid line.
Says parallel lines eventually meet "if you go far enough."	Parallel lines have the same slope and will never intersect, no matter how far out — that's the definition of parallel. The algebra confirms it with 0 = k (k ≠ 0).

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Scope flag

This outline stays within Objective 4 (systems part). Three-variable systems, matrices, and nonlinear systems are out of scope here (they appear in College Algebra 2e Chapter 7 but are beyond this week's objective). The inequalities treatment stays at the two-inequality, linear-boundary level; nonlinear boundaries are out of scope.