Chapter 3
System of Linear Equations
3.1 Linear Equations in
Two Variables

Forms of Linear Equation
 ax + by = c
 y = mx + b
(a, b, c are constants)
(b = y-intercept,
m = slope)
Solution of
System of Linear Equations

Given:
a) x + y = 7
b) x - y = -1


What are the coordinate
values such that both
equations are satisfied?
Answer: (3, 4)
x – y = -1
x+y=7
Solution: Substitution Method
Given:
y = -2x + 4
7x – 2y = 3
 Solve the system of equations
7x – 2(-2x + 4) = 3
7x + 4x – 8 = 3
11 x = 11
x=1

y = -2(1) + 4 = 2
 Solution
(1, 2)
 Check solution
Solution: Substitution Method
Given:
5x + 2y = 1
x – 3y = 7
 Solve the system of equations
x = 3y + 7
5(3y + 7) + 2y = 1
15y + 35 + 2y = 1
17y = -34
y = -2

x = 3(-2) + 7 = 1
 Solution
(1, -2)
 Check solution
Solution: Eliminating One Variable


Given:
3x + 4y = -10
5x – 2y = 18
Solve the system of equations
3x + 4y = -10
2(5x – 2y) = 2(18)
3x + 4y = -10
10x – 4y = 36
13x
x

= 26
=2
3(2) + 4y = -10
4y = -16
y = -4
Solution
(2, -4)
Lines with No Solutions

Given
3x – 2y = 6
6x – 4y = 18
-2(3x – 2y) = -2(6)
-6x + 4y = 12
6x – 4y = 18
0
=6
 Are there values of (x, y) for which 0 = 6?
Lines with No Solutions

Given
3x – 2y = 6
6x – 4y = 18
y = (2/3)x - 3
y = (2/3)x – (9/2)
a) x - 3(2x - 9) = 8
x – 6x + 27 = 8
-5x = -19
x = 19/5
y = 2(19/5) – 9
 Solve by substitution
= 38/5 – 9
method
= (38 – 45)/5
= -7/5
a) x – 3y = 8
Solution: (19/5, -7/5)
y = 2x - 9
b) 2x + 3y = 6
 Solve by elimination
2x – 3y = 6
Your Turn
method
b) 2x + 3y = 6
2x – 3y = 6
4x = 12
x=3
2(3) + 3y = 6
3y = 0
y=0
Solution: (3, 0)
Your Turn

Solve by elimination
method
2x – 5y = 13
5x + 3y = 17
3(2x – 5y) = 3(13)
5(5x + 3y) = 5(17)
6x – 15y = 39
25x + 15y = 85
-------------------32x = 124
x=4
6(4) – 15y = 39
24 – 15y = 39
-15y = 15
y = -1
Solution: (4, -1)
3.3 Systems of Linear Equations
in Three Variables

{(x, y, z) | ax + by + cz = d}
y-axis
(x, y, z)
x-axis
z-axis
Graph of
Linear Equation in 3 Variables

{(x, y, z) | 2x + 3y + z = 1}
Set of all points (x, y, z) in a particular
plane in 3-D.
y-axis
B(0, 1/3, 0)
A(1/2, 0, 0)
z-axis
C(0, 0, 1)
x-axis
X-Y Plane
y-axis
x-axis
{(x, y, z) | z = 0}
z-axis
X-Z Plane
y-axis
{(x, y, z) | y = 0}
x-axis
z-axis
Y-Z Plane
y-axis
{(x, y, z) | x = 0}
x-axis
z-axis
Systems of Linear Equations in
3 Variables

Given:
 air fare + hotel + car for $210
 hotel + car for $112
 air fare + hotel for $180

What is the cost of air fare only, hotel
only, or car only?
Solution
Let x = air fare
y = hotel
z = car
 x + y + z = 210
y + z = 112
x + y = 180
 Solution:
(x, y, z) = (98, 82, 30)

Solving Equations in 3 Variables (1)
Given:
5x – 2y – 4z = 3
3x + 3y + 2z = -3
-2x + 5y + 3z = 3
5x–3y–4z=3
3x+3y+2z=-3
5x – 2y – 4z = 3
2(3x + 3y + 2z) = 2(-3)
5x – 2y – 4z = 3
6x + 6y + 4z = -6
11x + 4y
= -3
3x+3y+2z=-3
-2x+5y+3z=3
3(3x + 3y + 2z) = 3(-3)
-2(-2x + 5y + 3z) = -2(3)
Solving Equations in 3 Variables (2)
11x + 4y
3x+3y+2z=-3
-2x+5y+3z=3
3(3x + 3y + 2z) = 3(-3)
-2(-2x + 5y + 3z) = -2(3)
9x + 9y + 6z
4x - 10y - 6z
13x y
11x+4y=-3
13x- y=-15
= -3
= -9
= -6
= -15
11x + 4y = -3
4(13x – y) = 4(-15)
11x + 4y = -3
52x – 4y = -60
63x
= -63
Solving Equations in 3 Variables (3)
63x = -63
x = -1
We had:
11x + 4y = -3
11(-1) + 4y = -3
4y = -3 + 11 = 8
y = 2
We had: 3x + 3y + 2z = -3
3(-1) + 3(2) + 2z = -3
-3 + 6 + 2z = -3
2z = -6
z = -3
Solution: (-1, 2, -3)
Check the solution
Solving Equations in 3 Variables (4)
Given: x +
z = 8
x + y + 2z = 17
x + 2y + z = 16
x+y+2z = 17
x+2y+z = 16
-2(x + y + 2z) = -2(17)
x + 2y + z) = 16
-2x - 2y - 4z = -34
x + 2y + z = 16
-x
- 3z = -18
x
-x
+ z
- 3z
2z
z =
= 8
= 18
= 10
5
Solving Equations in 3 Variables (5)
z = 5
We had: x +
z = 8
x + (5) = 8
x = 3
We had: x + y + 2z = 17
(3) + y + 2(5) = 17
3 + y + 10 = 17
y = 4
Solution: (3, 4, 5)
Check your solution
Incosistent Systems
Given: 2x + 5y +
z = 12
x – 2y + 4z = -10
-3x + 6y – 12z = 20
x-2y+4z=10
-3x+6y-12z=20
3(x – 2y + 4z) = 3(-10)
-3x + 6y – 12z = 20
3x – 6y + 12z = -30
-3x + 6y – 12z = 20
0 = -10 ???
•No (x, y, z) to satisfy this condition.
•No common point of intersection
•System of equations is inconsistent
Solution to air-hotel-car
problems
Let x = air fare
y = hotel
z = car
 a) x + y + z = 210
b) y + z = 112
c) x + y = 180

• a) x + y + z = 210
b) y + z = 112
c) x + y = 180
• Use a) and b) to eliminate z
x + y + z = 210
-1(
y + z) = -(112)
x + y + z = 210
- y - z = -112
-----------------------d) x = 98
• a) x + y + z = 210
b) y + z = 112
c) x + y = 180
• Use b) and c) to eliminate y
y + z = 112
-(x + y) = -(180)
y + z = 112
-x – y
= -180
-----------------------e) -x + z
= -68
• a) x + y + z = 210
b) y + z = 112
c) x + y = 180
• Use d) and e) to find z
d) x = 98
e) –x + z = -68
-(98) + z = -68
z = -68 + 98
z = 30
• a) x + y + z = 210
b) y + z = 112
c) x + y = 180
• Use c) and d) to find y
c) x + y = 180
d) x = 98
98 + y = 180
y = 180 – 98 = 82
Solution: (98, 82, 30) … (x: air, y: hotel, z: car)
Your Turn
Solve the following system
x + y + 2z = 11
x + y + 3z = 14
x + 2y – z = 5
 Is it consistent or inconsistent

Your Turn
a) x + y + 2z = 11
b) x + y + 3z = 14
c) x + 2y – z = 5
To rid of x
a) –(x + y + 2z) = -(11)
b)
x + y + 3z = 14
a) x + y + 2(3) = 11
b) x + y + 3(3) = 14
5 = 9
-x – y – 2z = -11
x + y + 3z = 14
----------------z = 3
x + y = 5
x + y = 9
No values of x, y will make this true.
Thus, no solution. (Inconsistent system)
3.4 Matrix Solution to
Linear Systems

Matrix
 Arrangement of number in rows and columns

Population (in mil.)
2000 2002 2004 2006
men
52
54
55
57
women
53
54
54
57
Augmented Matrix


4x + 4y
= 19
2y + 3z = 8
4x
- 5z = 7
4 4 0
0 2 3
4 0 -5
19
8
7
1
0
0
a
1
0
b
d
1
c
e
f
Matrix Row Operations
4x + 3y = -15
x + 2y = -1
Row-echelon form
4
1
3
2
-15
-1
1
0
•Interchange i th and j th rows: Ri
a
1
b
c
Rj
•Multiply each element in the ith row by k: kRi
•Add k times the elements in row i to
corresponding elements in Row j: kRi + Rj
Matrix Row Operations
Given a 2 x 3 matrix:
4
1
3
2
-15
-1
Interchange R1 and R2: R1
1
4
2
3
R2
-1
-15
Multiply each element in the 1st row by 3: 3R1
3(4) 3(3) 3(-15)
12 9
=
1
2
-1
1 2
-45
-1
Matrix Row Operations
4
1
3
2
-15
-1
Add 3 times the elements in row 1 to corresponding
elements in Row 2: 3R1 + R2
4
3
3(4)+1 3(3)+2
4 3 -15
=
13 11 -46
-15
3(-15)-1
Your Turn
Given:
3
4
-2
-3
5
1
Perform the following operation: R1
4
3
-3 1
-2 5
R2
Your Turn
Given:
3
4
-2 5
-3 1
Perform the following operation: 2R2
3
-2
5
2(4) 2(-3) 2(1) =
3 -2
8 -6
5
2
Your Turn
3
Given:
4
-2 5
-3 1
Perform the following operation: 3R1 + R2
3
-2
5
3(3)+4 3(-2)-3 3(5)+1
3 -2 5
=
13 -9 16
Matrix Row Operations
4x + 4y
= 19
2y + 3z = 8
4x
- 5z = 7
R1
R2
kR2 (k = -2)
kR1+R3 (k = -1)
0
4
4
2 3
4 0
0 -5
4
0
4
4 0
2 3
0 -5
8
19
7
0 2 3
-8 -8 0
4 0 -5
0 2 3
-8 -8 0
4 -2 -8
8
-38
7
8
-38
-1
19
8
7
Solving Linear System in 2 Variables
4x – 3y = -15
x + 2y = -1
4 -3
1 2
-15
-1
Recall the strategy.
4
1
3
2
-15
-1
1
0
a
1
Then, the last row says: y = c
The first row says:
x + ay = b
b
c
Solving thy Linear System in 2 Variables
4x – 3y = -15
x + 2y = -1
R2 (to make e11 = 1)
R1
4 -3
1 2
(-1/11)R2
1
0
4 -3
1 2
2
-11
-15
-1
-4R1 + R2 (to make e21 = 0)
-15
-1
1 2
4 -3
-1
-15
2
1
-1
1
(to make e12 = 1)
-1
-11
1
0
Solving System in 2 Variables
1
0
2
1
-1
1
y = 1
x + 2(1) = -1
x = -3
Check:
4x – 3y
= -15
4(-3) – 3(1) =
-12 – 3
= -15
x + 2y
= -1
Solution: (-3, 1) (-3) + 2(1) =
-3 + 2
= -1
Your Turn

Solve the following using matrix method.
3x – 6y = 1
2x – 4y = 2/3
3 -6
2 -4
1
2/3
1 a
0 1
b
c
(1/3)R1 (to make e11 = 1)
(1/3)3 (1/3(-6) (1/3)1
2
-4
2/3
=
1 -2 1/3
2 -4 2/3
Your Turn
(1/3)3 (1/3(-6) (1/3)1
2
-4
2/3
=
1 -2 1/3
2 -4 2/3
2R1-R2 (to make e21 = 0)
1
2(1)-2
-2
1/3
=
2(-2)-4 2(1/3) – 2/3
1 -2 1/3
0 -8 0
(-1/8)R2 (to make e22 = 1)
1
(-1/8)0
-2
1/3
(-1/8)(-8) (-1/8)0
=
1 -2 1/3
0 1 0
Your Turn
Check:
1 -2 1/3
0 1 0
y=0
x – 2y = 1/3
x – 0 = 1/3
x = 1/3
Solution: (1/3, 0)
3x – 6y = 1
3(1/3) – 6(0) = 1
1=1
2x – 4y = 2/3
2(1/3) – 4(0) = 2/3
2/3 – 0 = 2/3
2/3 = 2/3
Your Turn

Solve the following using matrix method.
-3x + 4y = 12
2x + y = 3
Solving System in 3 Variables
x + y + z = 210
y + z = 112
x + y
= 180
(-1)R1+R3
1
0
0
(-1)R3
x + y + z = 210
y + z = 112
z = 30
1
0
1
1 1
1 1
0 -1
1
0
0
1
1
0
1
1
1
1
1
0
210
112
180
210
112
-30
1
1
1
210
112
30
z = 30
y + 30 = 112 → y = 82
x + 82 + 30 = 210 → z = 98
Your Turn

Solve the following system of equations
 3x + y + 2z = 31
x + y + 2z = 19
x + 3y + 2z = 25
3.5:

2 x 2 Matrix
a1
a2

Determinants
b1
b2
Determinant
a1
a2
b1
b2
= a1b2 – a2b1
Determinants
2 4
-3 -5
= 2(-5) – (-3)4 = -10 + 12 = 2
10
6
9
5
= ?
4 3
-5 -8
= ?
Cramer’s Rule
Given:
a1x + b1y = c1
a2x + b2y = c2
x =
Dx
x = ------D
c1
c2
b1
b2
a1
a2
b1
b2
y =
Dy
y = ------D
a1
a2
c1
c2
a1
a2
b1
b2
Cramer’s Rule
Given:
5x - 4y = 2
6x - 5y = 1
2
1
-4
-5
5
6
-4
-5
5
6
2
1
5
6
-4
-5
x =
y =
-10 – (-4)
-6
= ------------ = ---- = 6
-25 – (-24)
-1
5 – 12
-7
= -------- = ---- = 7
-1
-1
(6, 7)
Your Turn

Use Cramer’s rule to solve
1. 12x + 3y = 15
2x – 3y = 13
2. x – 3y = 4
3x – 4y = 12
Solution for 1
12x + 3y = 15
2x – 3y = 13
D
12 3
= 2 -3
15 3
Dx = 12 -3
= 12(-3) – 2(3) = -42
= 15(-3) – 3(12) = -84
Dx
-84
x = ---- = ----- = 2
D
-42
Solution for 1
12x + 3y = 15
2x – 3y = 13
D
12 3
= 2 -3
12 15
Dy = 2 13
= 12(-3) – 2(3) = -42
= 12(13) – 2(15) = 126
Dy
126
y = ---- = ----- = -3
D
-42
Solution: (2, -3)
Solution for 2
x
3x
D
- 36 = 4
– 3y = 12
=
1 -3
3 -3
4 -3
Dx = 12 -3
= 1(-3) – 3(-3) = 6
= 4(-3) – 12(-3) = 24
Dx
24
y = ---- = ---- = 4
D
6
Solution for 2
x
3x
- 36 = 4
– 3y = 12
=
1 -3
3 -3
= 1(-3) – 3(-3) = 6
Dy =
1 4
3 12
= 1(12) – 3(4) = 0
D
Dy
0
y = ---- = ---- = 0
D
6
Solution: (4, 0)
Determinant of a 3 x 3 Matrix
a1 b1 c1
b2 c2
b1 c1
b1 c1
a2 b2 c2 = a1 b3 c3 – a2 b3 c3 + a3 b2 c2
a3 b3 c3
E.g.
2 3 4
5 2 1
4 3 6
= ?
Determinant of a 3 x 3 Matrix
Given: a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Dx
x = ---D
a1 b1 c1
D = a2 b2 c2
a3 b3 c3
Dx
Dy
y = ---D
d1 b2 c3
= d2 b2 c3
d3 b3 c3
a1 d1 c1
Dy = a2 d2 c2
a3 d3 c3
Dz =
a1 b1 d1
a2 b2 d2
a3 b3 d3
Dz
z = ---D
Example
Given: x + 2y - z = -4
x + 4y - 2z = -6
2x + 3y + z = 3
1
D = 1
2
Dx
2 -1
4 -2
3 1
-4 2 -1
= -6 4 -2
3 3 1
1 -4 -1
Dy = 1 -6 -2
2 3 1
Dz =
1
1
2
2 -4
4 -6
3 3
Example
1
D = 1
2
2 -1
4 -2
4 -2 = 1 3 1
3 1
2 -1
2 -1
- 1 3 1 + 2 4 -2
= (4 – (-6)) – (2 – (-3)) + 2(-4 – (-4))
= 10 - 5 - 0 = 5
Dx
-4 2 -1
4 -2
2 -1
2 -1
= -6 4 -2 = -4 3 1 - (-6) 3 1 + 3 4 -2
3 3 1
= (-4)(4 + 6) – (-6)(2 + 3) + (3)(-4 + 4)
= -40 + 30 + 0 = 10
-10
x = ----- = -2
5
Your Turn

Calculate
1. y = Dy / D
2. Z = Dz / D