Study Guide - Module 5 - College Algebra

1. Systems of Linear Equations

1.1 Substitution Method

Use substitution when one equation is already solved for a variable or can be easily solved for a variable.

Solve one equation for one variable (if not already done)
Substitute that expression into the other equation
Solve the resulting single-variable equation
Substitute back to find the other variable
Check your solution in both original equations

Example: Solve y = 2x - 3 and x + y = 9

Step 1: First equation already solved for y

Step 2: Substitute into second: x + (2x - 3) = 9

Step 3: Solve: 3x - 3 = 9 → 3x = 12 → x = 4

Step 4: Back-substitute: y = 2(4) - 3 = 5

Solution: (4, 5)

1.2 Elimination (Addition) Method

Use elimination when coefficients of one variable are opposites or can be made opposites by multiplication.

Multiply one or both equations to make coefficients of one variable opposites
Add the equations to eliminate that variable
Solve the resulting single-variable equation
Substitute back into either original equation
Check your solution in both original equations

Example: Solve 3x + 2y = 16 and 5x - 2y = 8

Step 1: Coefficients of y are already opposites (2 and -2)

Step 2: Add equations: 8x = 24

Step 3: Solve: x = 3

Step 4: Substitute: 3(3) + 2y = 16 → 2y = 7 → y = 3.5

Solution: (3, 3.5)

1.3 Types of Solutions

Type	Number of Solutions	Graphical Interpretation	Algebraic Result
Consistent Independent	One unique solution (x, y)	Lines intersect at one point	One solution found
Consistent Dependent	Infinitely many solutions	Lines are the same (coincident)	True statement (0 = 0)
Inconsistent	No solution	Parallel lines (never intersect)	False statement (0 = 5)

2. Matrices and Gaussian Elimination

2.1 Augmented Matrix

An augmented matrix represents a system of equations by writing coefficients and constants in matrix form.

System: ax + by = e
cx + dy = f

Augmented Matrix: [a b | e]
[c d | f]

Example: Write the augmented matrix for 2x - 3y = 7 and x + y = 5

Answer: [2 -3 | 7]
[1 1 | 5]

2.2 Row Operations

Three types of elementary row operations:

Swap rows: Ri ↔ Rj
Multiply a row by a nonzero constant: kRi → Ri
Add a multiple of one row to another: Ri + kRj → Ri

2.3 Gaussian Elimination

Write the augmented matrix
Use row operations to get zeros below each pivot (leading entry)
Continue until the matrix is in row-echelon form (triangular shape)
Use back-substitution to find the values of the variables

Row-Echelon Form: Each row's first nonzero entry (pivot) is 1, and all entries below each pivot are 0.

2.4 Gauss-Jordan Elimination

Gauss-Jordan continues beyond row-echelon form to create zeros above each pivot as well, resulting in reduced row-echelon form where the solution can be read directly.

3. Nonlinear Systems

3.1 Types of Nonlinear Systems

A nonlinear system includes at least one equation that is not linear (contains x², y², xy, etc.)

Common combinations:

Line and parabola (0, 1, or 2 solutions)
Line and circle (0, 1, or 2 solutions)
Two parabolas (0, 1, 2, 3, or 4 solutions)
Circle and ellipse (0, 1, 2, 3, or 4 solutions)

3.2 Substitution Method for Nonlinear Systems

Substitution is the preferred method for nonlinear systems. Solve the simpler (usually linear) equation for one variable and substitute into the nonlinear equation.

Example: Solve y = x² and y = 2x + 3

Step 1: Set equations equal: x² = 2x + 3

Step 2: Rearrange: x² - 2x - 3 = 0

Step 3: Factor: (x - 3)(x + 1) = 0

Step 4: Solve: x = 3 or x = -1

Step 5: Find y-values: When x = 3, y = 9; when x = -1, y = 1

Solutions: (3, 9) and (-1, 1)

3.3 Elimination for Nonlinear Systems

Elimination can work when both equations have the same type of terms (both have x² and y²)

Example: Solve x² + y² = 25 and x² - y² = 7

Add equations: 2x² = 32 → x² = 16 → x = ±4

Substitute into first: 16 + y² = 25 → y² = 9 → y = ±3

Solutions: (4, 3), (4, -3), (-4, 3), (-4, -3)

4. Systems of Inequalities and Linear Programming

4.1 Graphing Linear Inequalities

Graph the boundary line (solid for ≤ or ≥, dashed for < or >)
Choose a test point not on the line (often (0, 0) if not on the line)
Substitute the test point into the inequality
If true, shade the side containing the test point; if false, shade the opposite side

Quick Rules:
• y > mx + b → shade above the line
• y < mx + b → shade below the line
• Use solid line for ≤ or ≥ (equality included)
• Use dashed line for < or > (equality not included)

4.2 Systems of Inequalities

The solution to a system of inequalities is the feasible region where all inequalities are satisfied simultaneously (the overlapping shaded region).

To graph a system:

Graph each inequality on the same coordinate plane
Identify the region where all shaded areas overlap
The vertices (corner points) are where boundary lines intersect

4.3 Linear Programming

Corner Point Theorem: If a linear objective function has a maximum or minimum value, it will occur at one or more vertices of the feasible region.

Steps to solve a linear programming problem:

Define variables for the quantities to be determined
Write the objective function (what you want to maximize or minimize)
Write all constraints as inequalities
Graph the feasible region and find the vertices
Evaluate the objective function at each vertex
The vertex with the largest value gives the maximum; smallest gives the minimum

Example: Maximize P = 3x + 4y subject to:

x + y ≤ 10
2x + y ≤ 16
x ≥ 0, y ≥ 0

Vertices: (0, 0), (0, 10), (6, 4), (8, 0)

Evaluate P:
At (0, 0): P = 0
At (0, 10): P = 40
At (6, 4): P = 3(6) + 4(4) = 34
At (8, 0): P = 24

Maximum: P = 40 at (0, 10)

4.4 Setting Up Word Problems

Common constraint types:

Resource constraints: Total available ≥ amount used (time, materials, money)
Minimum requirements: Variable ≥ minimum value
Maximum capacities: Variable ≤ maximum value
Non-negativity: x ≥ 0, y ≥ 0 (cannot produce negative quantities)

5. Practice Problems with Solutions

Problem 1: Substitution

Solve: y = x + 4 and 2x - y = 1

Solution:
Substitute y = x + 4 into second equation:
2x - (x + 4) = 1
2x - x - 4 = 1
x = 5
Then y = 5 + 4 = 9
Answer: (5, 9)

Problem 2: Elimination

Solve: 4x + 3y = 10 and 2x - 3y = 8

Solution:
Add equations (y-coefficients are opposites):
6x = 18 → x = 3
Substitute: 4(3) + 3y = 10 → 3y = -2 → y = -2/3
Answer: (3, -2/3)

Problem 3: Inconsistent System

Solve: x - 2y = 3 and 2x - 4y = 10

Solution:
Multiply first equation by -2: -2x + 4y = -6
Add to second: 0 = 4 (False!)
Answer: No solution (inconsistent)

Problem 4: Dependent System

Solve: 2x + y = 6 and 4x + 2y = 12

Solution:
Multiply first equation by -2: -4x - 2y = -12
Add to second: 0 = 0 (Always true!)
Answer: Infinitely many solutions; y = -2x + 6

Problem 5: Nonlinear System

Solve: y = x² - 2 and y = x

Solution:
Set equal: x² - 2 = x
x² - x - 2 = 0
(x - 2)(x + 1) = 0 → x = 2 or x = -1
When x = 2: y = 2; when x = -1: y = -1
Answer: (2, 2) and (-1, -1)

Problem 6: System of Inequalities

Graph and find vertices: x + y ≤ 5, x - y ≤ 1, x ≥ 0, y ≥ 0

Solution:
Boundary lines: x + y = 5, x - y = 1, x = 0, y = 0
Vertices:
• (0, 0): x = 0 and y = 0
• (0, 5): x = 0 and x + y = 5
• (3, 2): x + y = 5 and x - y = 1 (solve: 2x = 6, x = 3, y = 2)
• (1, 0): x - y = 1 and y = 0
Vertices: (0, 0), (0, 5), (3, 2), (1, 0)

Problem 7: Linear Programming

Minimize C = 2x + 3y subject to: x + y ≥ 4, 2x + y ≥ 6, x ≥ 0, y ≥ 0

Solution:
Vertices of feasible region:
• (0, 6): x = 0 and 2x + y = 6
• (2, 2): x + y = 4 and 2x + y = 6
• (4, 0): x + y = 4 and y = 0
Evaluate C:
At (0, 6): C = 18
At (2, 2): C = 10
At (4, 0): C = 8
Minimum: C = 8 at (4, 0)

Problem 8: Word Problem

Problem: A store sells small and large gift baskets. Small baskets cost $20 and large cost $35. The store needs to make at least $700 and can make at most 30 baskets. How many of each should be made to minimize cost?

Solution:
Let x = small baskets, y = large baskets
Constraints: 20x + 35y ≥ 700, x + y ≤ 30, x ≥ 0, y ≥ 0
Objective: Minimize total baskets = x + y
This would require graphing and finding vertices.
Key vertices would be found at intersections of constraint lines.

6. Test-Taking Tips

Always check your solutions by substituting back into the original equations
For word problems, define your variables clearly and write units
When graphing inequalities, remember that solid lines include the boundary (≤, ≥) and dashed lines do not (<, >)
For linear programming, always evaluate the objective function at ALL vertices
Watch for inconsistent and dependent systems - they require special answers
Use elimination when coefficients are already opposites or easy to make opposites
Use substitution when a variable is already isolated or easy to isolate
Show your work - partial credit is often awarded for correct process

Module 5 Study Guide