Module 5 Quick Reference
Systems of Equations and Inequalities - One-Page Cheat Sheet
Methods for Solving Linear Systems
Substitution Method
When to use: One variable is already isolated, or easy to isolate
Steps:
- Solve one equation for one variable
- Substitute into the other equation
- Solve for remaining variable
- Back-substitute to find other variable
Elimination Method
When to use: Coefficients are (or can be made) opposites
Steps:
- Multiply to make coefficients opposites
- Add equations to eliminate variable
- Solve for remaining variable
- Substitute back to find other variable
Types of Solutions
| Type | Solutions | Graph | Algebraic Sign |
|---|---|---|---|
| Consistent Independent | One (x, y) | Lines intersect | One solution |
| Consistent Dependent | Infinitely many | Same line | 0 = 0 (true) |
| Inconsistent | None | Parallel lines | 0 = k (false) |
Matrix Methods
Augmented Matrix
System: ax + by = e → [a b | e]
cx + dy = f → [c d | f]
cx + dy = f → [c d | f]
Elementary Row Operations
- Swap rows: Ri ↔ Rj
- Multiply row: kRi → Ri (k ≠ 0)
- Add multiple of row: Ri + kRj → Ri
Row-Echelon Form
- Pivots (leading 1s) form stairs
- Zeros below each pivot
- Use back-substitution
Reduced Row-Echelon
- Zeros above AND below pivots
- Each pivot is only nonzero in column
- Solution can be read directly
Nonlinear Systems
Key Strategy: Use substitution for nonlinear systems. Solve the simpler equation for one variable and substitute into the other.
Possible Number of Solutions
- Line and parabola: 0, 1, or 2 solutions
- Line and circle: 0, 1, or 2 solutions
- Two quadratics: 0, 1, 2, 3, or 4 solutions
Common Technique
Example: y = x2 and y = 2x + 3
Set equal: x2 = 2x + 3
Rearrange: x2 - 2x - 3 = 0
Solve quadratic: (x - 3)(x + 1) = 0
Find y for each x value
Set equal: x2 = 2x + 3
Rearrange: x2 - 2x - 3 = 0
Solve quadratic: (x - 3)(x + 1) = 0
Find y for each x value
Graphing Inequalities
Boundary Line
- Solid line: ≤ or ≥ (equality included)
- Dashed line: < or > (equality not included)
Shading Direction
Test Point Method:
- Pick test point (usually (0,0))
- Substitute into inequality
- If true, shade that side
- If false, shade other side
Quick Rules:
- y > mx + b → shade above
- y < mx + b → shade below
- y ≥ mx + b → shade above (solid)
- y ≤ mx + b → shade below (solid)
Systems of Inequalities
Feasible Region: The overlapping area where all inequalities are satisfied. Vertices occur where boundary lines intersect.
Linear Programming
Corner Point Theorem
If an objective function has a max or min value, it occurs at a vertex (corner point) of the feasible region.
Steps to Solve
- Define variables: x and y represent quantities
- Objective function: What to maximize or minimize (e.g., P = ax + by)
- Write constraints: All restrictions as inequalities
- Graph feasible region: Find where all constraints overlap
- Find vertices: Points where boundary lines intersect
- Evaluate objective: Calculate P at each vertex
- Identify optimal: Largest P for max, smallest for min
Common Constraint Types
| Type | Example | Inequality |
|---|---|---|
| Resource limit | 60 hours available | 2x + 3y ≤ 60 |
| Minimum requirement | At least 10 units | x + y ≥ 10 |
| Non-negativity | Cannot be negative | x ≥ 0, y ≥ 0 |
Example Template
Given: Product A makes $10, Product B makes $15
Machine time: 2A + 3B ≤ 60 hours
Materials: 4A + 2B ≤ 80 units
Variables: Let x = units of A, y = units of B
Objective: Maximize P = 10x + 15y
Constraints: 2x + 3y ≤ 60, 4x + 2y ≤ 80, x ≥ 0, y ≥ 0
Solution: Graph, find vertices, evaluate P at each
Machine time: 2A + 3B ≤ 60 hours
Materials: 4A + 2B ≤ 80 units
Variables: Let x = units of A, y = units of B
Objective: Maximize P = 10x + 15y
Constraints: 2x + 3y ≤ 60, 4x + 2y ≤ 80, x ≥ 0, y ≥ 0
Solution: Graph, find vertices, evaluate P at each
Quick Formula Summary
Substitution: Solve for one variable, substitute into other equation
Elimination: aRi + bRj → Ri (make coefficients opposites, add)
Augmented Matrix: [coefficients | constants]
Nonlinear: Set equal, solve resulting equation
Feasible Region: All constraints satisfied (overlapping shaded area)
Linear Programming: Evaluate objective P at all vertices, pick max/min
Elimination: aRi + bRj → Ri (make coefficients opposites, add)
Augmented Matrix: [coefficients | constants]
Nonlinear: Set equal, solve resulting equation
Feasible Region: All constraints satisfied (overlapping shaded area)
Linear Programming: Evaluate objective P at all vertices, pick max/min
Essential Tips
- Always check solutions in original equations
- Watch for 0 = 0 (infinitely many) vs 0 = k (no solution)
- Solid line (≤, ≥) includes boundary; dashed (<, >) does not
- Evaluate objective function at ALL vertices in linear programming
- Non-negativity constraints (x ≥ 0, y ≥ 0) are almost always required in word problems
- For nonlinear systems, prefer substitution over elimination