Systems of Linear Equations

Review: Systems of Two Equations

System: Multiple equations with same variables

Two equations, two variables:

2x + y = 5
x - y = 1

Methods: Graphing, substitution, elimination

Solution: Point where lines intersect

Three Variables, Three Equations

General form:

ax + by + cz = d
ex + fy + gz = h
ix + jy + kz = l

Solution: Ordered triple (x, y, z)

Geometric meaning: Point where three planes intersect

Example: Simple System

System:

x + y + z = 6
x + y - z = 2
x - y + z = 4

Solution: (3, 1, 2)

Check:

• 3 + 1 + 2 = 6 ✓
• 3 + 1 - 2 = 2 ✓
• 3 - 1 + 2 = 4 ✓

Method 1: Elimination

Strategy: Eliminate one variable at a time

Steps:

1. Eliminate one variable from two pairs of equations
2. Solve resulting 2×2 system
3. Back-substitute to find third variable

Example: Elimination Method

System:

2x + y - z = 3    (1)
x - y + 2z = 1    (2)
3x + 2y + z = 10  (3)

Step 1: Eliminate y from (1) and (2)

Add equations (1) and (2):

2x + y - z = 3
x - y + 2z = 1
--------------
3x + z = 4        (4)

Step 2: Eliminate y from (2) and (3)

Multiply (2) by 2:

2x - 2y + 4z = 2

Add to (3):

2x - 2y + 4z = 2
3x + 2y + z = 10
----------------
5x + 5z = 12      (5)

Step 3: Solve system of (4) and (5)

From (4): 3x + z = 4 → z = 4 - 3x

Substitute into (5):

5x + 5(4 - 3x) = 12
5x + 20 - 15x = 12
-10x = -8
x = 0.8

Step 4: Find z

z = 4 - 3(0.8) = 4 - 2.4 = 1.6

Step 5: Find y using equation (1)

2(0.8) + y - 1.6 = 3
1.6 + y - 1.6 = 3
y = 3

Solution: (0.8, 3, 1.6)

Method 2: Substitution

Strategy: Solve for one variable, substitute into others

Best when: One equation already solved for a variable

Example: Substitution Method

System:

z = 2x + 1        (1)
x + y + z = 8     (2)
2x - y + z = 5    (3)

Step 1: Substitute (1) into (2)

x + y + (2x + 1) = 8
3x + y = 7
y = 7 - 3x        (4)

Step 2: Substitute (1) and (4) into (3)

2x - (7 - 3x) + (2x + 1) = 5
2x - 7 + 3x + 2x + 1 = 5
7x - 6 = 5
7x = 11
x = 11/7

Step 3: Find y and z

y = 7 - 3(11/7) = 7 - 33/7 = 49/7 - 33/7 = 16/7
z = 2(11/7) + 1 = 22/7 + 7/7 = 29/7

Solution: (11/7, 16/7, 29/7)

Matrix Method (Introduction)

System can be written as matrix equation AX = B

Example:

2x + y - z = 3
x - y + 2z = 1
3x + 2y + z = 10

Matrix form:

[2   1  -1] [x]   [3]
[1  -1   2] [y] = [1]
[3   2   1] [z]   [10]

Solve using: Row operations, inverse matrix, or calculator

Gaussian Elimination

Systematic elimination using augmented matrix

Augmented matrix: Coefficients and constants

[2   1  -1 | 3]
[1  -1   2 | 1]
[3   2   1 | 10]

Goal: Reduce to row echelon form

Example: Gaussian Elimination

System:

x + y + z = 6
2x - y + z = 3
x + 2y - z = 0

Augmented matrix:

[1   1   1 | 6]
[2  -1   1 | 3]
[1   2  -1 | 0]

Row operations:

R2 - 2R1 → R2:

[1   1   1 | 6]
[0  -3  -1 | -9]
[1   2  -1 | 0]

R3 - R1 → R3:

[1   1   1 | 6]
[0  -3  -1 | -9]
[0   1  -2 | -6]

R2 ÷ (-3) → R2:

[1   1   1 | 6]
[0   1  1/3 | 3]
[0   1  -2 | -6]

R3 - R2 → R3:

[1   1   1 | 6]
[0   1  1/3 | 3]
[0   0 -7/3 | -9]

Back-substitute:

-7z/3 = -9 → z = 27/7
y + z/3 = 3 → y = 3 - 9/7 = 12/7
x + y + z = 6 → x = 6 - 12/7 - 27/7 = 3/7

Solution: (3/7, 12/7, 27/7)

Types of Solutions

One solution: Three planes intersect at one point (consistent, independent)

No solution: Planes don't share common point (inconsistent)

Infinitely many solutions: Planes intersect along line or coincide (consistent, dependent)

Example: No Solution

System:

x + y + z = 1
x + y + z = 2
2x + 2y + 2z = 5

First two equations contradictory: x + y + z cannot equal both 1 and 2

No solution (parallel planes)

Example: Infinite Solutions

System:

x + y + z = 4
2x + 2y + 2z = 8
3x + 3y + 3z = 12

All three equations equivalent (same plane)

Solution: All points on plane x + y + z = 4

Express as: z = t, y = s, x = 4 - s - t (parametric form)

Four or More Variables

Same methods apply

More elimination steps needed

Matrix methods become essential

Example: Four Variables

System:

w + x + y + z = 10
w + x - y - z = 0
w - x + y - z = 2
w - x - y + z = -2

Add pairs to eliminate variables systematically

Solution: (3, 2, 4, 1)

Applications: Real-World Problems

Chemistry: Balancing chemical equations

Business: Production planning, resource allocation

Nutrition: Meeting dietary requirements

Engineering: Network analysis, circuit design

Example: Investment Problem

Problem: Invest $10,000 in three accounts

Conditions:

• Total investment: $10,000
• First account pays 4%, second 6%, third 8%
• Want $600 annual interest
• Invest twice as much in first as second

Variables:

• x = amount in first account
• y = amount in second account
• z = amount in third account

Equations:

x + y + z = 10000        (total)
0.04x + 0.06y + 0.08z = 600   (interest)
x = 2y                   (condition)

Substitute x = 2y:

2y + y + z = 10000 → 3y + z = 10000
0.04(2y) + 0.06y + 0.08z = 600
0.08y + 0.06y + 0.08z = 600
0.14y + 0.08z = 600

Solve:

From first: z = 10000 - 3y
Substitute: 0.14y + 0.08(10000 - 3y) = 600
0.14y + 800 - 0.24y = 600
-0.10y = -200
y = 2000

Therefore:

x = 2(2000) = 4000
z = 10000 - 3(2000) = 4000

Answer: $4000 at 4%, $2000 at 6%, $4000 at 8%

Example: Mixture Problem

Problem: Chemist needs 100 liters of 40% acid solution

Available:

• Solution A: 20% acid
• Solution B: 50% acid
• Solution C: 80% acid

Use twice as much A as C

Equations:

a + b + c = 100           (total volume)
0.20a + 0.50b + 0.80c = 40    (acid content)
a = 2c                    (condition)

Solution: 40L of A, 40L of B, 20L of C

Solving with Technology

Graphing calculators: Matrix operations, rref function

Computer algebra systems: Mathematica, Maple, MATLAB

Spreadsheets: Excel solver

Programming: Python (NumPy), R

Benefits: Fast, accurate for large systems

Practice