Algebra 2

Chapter 3 Notes

Systems of Linear Equalities and Inequalities

A system of 2 linear equations in 2 variables, x & y consists of 2 equations of the following form:

A x + B y = CD x + E y = F

A, B, C, D, E and F all represent constant values.

A solution of a system of linear equations in 2 variables is an ordered pair (x,y) that satisfies both equations.

Example 1: Checking solutions of a linear system. Are ( 2 , 2 ) and ( 0 , − 1 ) solutions of the following system:

3 x – 2 y = 2 x + 2 y = 6

3 x – 2 y = 2 x + 2 y = 6 3 ( 2 ) – 2 ( 2 ) = 2 ( 2 ) + 2 ( 2 ) = 66 – 4 = 2 2 + 4 = 62 = 2√ 6 = 6 √

3 x – 2 y = 2 x + 2 y = 6 3 ( 0 ) – 2 (− 1 ) = 2 ( 0 ) + 2 (− 1 ) = 60 + 2 = 2 0 − 2 = 62 = 2√ − 2 ≠ 6

Solution works for both

Solution does NOT work for both

Solving Linear Systems by Graphing 3.1

x 2 x – 3 y = 1 y x x + y = 3 y

0 2 (0) – 3 y = 1 – 3 y = 1

− 1 3

0 0 + y = 3 3

1 2

2 x – 3 (0) = 1 0 3 x + 0 = 3 0

• ••

•

•

2 x – 3 y = 1

x + y = 3

( 2 , 1 )

2 x – 3 y = 12 ( 2 ) – 3 ( 1 ) = 14 – 3 = 11 = 1 √

x + y = 32 + 1 = 3 3 = 3 √

Solving Linear Systems by Graphing 3.1

Graphical Interpretation Algebraic Interpretations

The graph of the system is a pair of lines that intersect in 1 point

The system has exactly 1 solution

The graph of the system is a single line The system has infinitely many solutions

The graph of the system is a pair of parallel lines so that there is no point of intersection

The system has no solution

Exactly 1 solution Infinitely many solutions No solution

Number of Solutions of a Linear System

3.1

Substitution Method:

1. Solve for one equation

2. Substitute the expression from step 1 into the other equation, then solve for the other variable

3. Substitute the value from step 2 into the revised equation from step 1, then solve

Example 1: 3 x + 4 y = − 4 [1st equation ] x + 2 y = 2 [2nd equation ]

1) Solve for x in equation 2 x + 2 y = 2 − 2 y − 2 yx = − 2 y + 2

2) Substitute 3 x + 4 y = − 43 (− 2 y + 2 ) + 4 y = − 4− 6 y + 6 + 4 y = − 4 − 2 y + 6 = − 4 − 2 y = − 10

y = 5

3). Use value for y to get x: x = − 2 y + 2 x = − 2 ( 5 ) + 2 x = − 10 + 2 = − 8

Check: 3 x + 4 y = − 43 (− 8 ) + 4 (5 ) = − 4 − 24 + 20 = − 4 − 4 = − 4 √

Solving Linear System Algebraically 3.2

7 x − 12 y = − 22 − 5 x + 8 y = 14

2 [ 7 x − 12 y = − 22 ] 3 [− 5 x + 8 y = 14 ]

14 x − 24 y = − 44 − 15 x + 24 y = 42 − 1 x = − 2

x = 2− 5 x + 8 y = 14− 5 (2) + 8 y = 14 − 10 + 8 y = 14 8 y = 24 y = 3 ( x , y )

( 2, 3 )

Check: 7 x − 12 y = − 227 ( 2 ) − 12 ( 3 ) = − 2214 − 36 = − 22 − 22 = − 22 √

Solving Linear System Algebraically

Linear Combination Method:

3.2

2 x + 3 y = 5 x − 5 y = 9 x = 5 y + 9

2 ( 5 y + 9 ) + 3 y = 510 y + 18 + 3 y = 513 y + 18 = 513 y = − 13 y = − 1

Check:2 x + 3 y = 52 (4) + 3 ( − 1 ) = 58 − 3 = 55 = 5 √

x − 5 y = 9 x − 5 (− 1) = 9x + 5 = 9 x = 9 − 5 x = 4

3 x + 5 y = − 163 x − 2 y = − 9− 1 [ ]

3 x + 5 y = − 16 − 3 x + 2 y = + 9 7 y = − 7 y = − 1

3 x + 5 (− 1) = − 163 x − 5 = − 163 x = − 11

x = − 11 3Check:

3 x + 5 y = − 163 (− 11 ) + 5 (− 1) = − 16 3− 11 − 5 = − 16

− 16 = − 16 √

( x , y )(− 11 , − 1 )

3

Solving Linear System Algebraically

Linear Combination Method:

Substitution Method:

3.2

y ≥ ─ 3 x ─ 1

x y

0 −1

−1 3

0

Test ( 0 , 0 )y > ─ 3 x ─ 1 0 > ─ 3 ( 0 ) ─ 1

0 > ─ 1 √

y < x + 2

x y

0 2

− 2 0

Test ( 0 , 0 ) y < x + 2

0 < ( 0 ) + 20 < 2 √

••

y = ─ 3 x ─

1

•

•

y = x

+ 2

BLUE ONLY

BOTHRED & BLUE

NEITHER RED nor BLUE

A solution of a system of linear inequalities is an ordered pair that is a solution of each inequality in the system or a coordinate in BOTH solutions of each inequality

y < x

+ 2

y ≥ ─ 3 x ─

1

RED ONLY

Graphing and Solving of Linear Inequalities

3.3

x ≥ 0

y ≥ 0

4 x + 3 y ≤ 24 x ≥ 0

y ≥ 0

4 x + 3 y ≤ 24

•

•

•Test Point (2, 2)

2 ≥ 0 √

2 ≥ 0 √

4 (2) + 3 (2) ≤ 24 8 + 6 ≤ 24 14 ≤ 24 √

Graphing a System of 3 Inequalities 3.3

Real life problems involve a process called OPTIMIZATION = finding the maximum or minimum value of some quantity.

Linear Programming is the process of optimizing a linear objective function subject to a system of linear inequalities called constraints. The graph of the system of constraints is called the feasible region.

Optimal solution of a Linear Programming problemIf an objective function has a maximum or a minimum value, then it must occur at a vertex of the feasible region. If the objective function is bounded,, then it has both a maximum and a minimum value.

• •

••

••

•

Bounded Region Unbounded Region

Linear Programming (a type of optimization)

3.4

Optimum value, maximum or minimum, occur at vertices of a feasible region.

Maximum = (3.6) or C = 2 x + 4 y “R” C = 2 (3) + 4 (6)

C = 6 + 24 = 30

Minimum = (0.0) or C = 2 x + 4 y “O” C = 2 (0) + 4 (0)

C = 0 + 0 = 0

••

••

C = 2 x + 4 y

(0,5)

(0,0) (6,0)

(3,6)

O

PR

V

(0,5)“P” C = 2 (0) + 4 (5) C = 0 + 20 = 20

(6,0)“V” C = 2 (6) + 4 (0) C = 12 + 0 = 12

Optimum Value of a Feasible Region

feasible region

3.4

•

••(8,0)

(0,8)

(0,0)

Objective Function: C = 3 x + 4 y

x ≥ 0

Contraints: y ≥ 0

x + y ≤ 8 {

Vertices: C = 3 x + 4 y

(0,0) C = 3 (0) + 4 (0) = 0

(8,0) C = 3 (8) + 4 (0) = 24

(0,8) C = 3 (0) + 4 (8) = 32

minimum value

maximum value

Find Maximum and Minimum Values

Ex 1

3.4

•

•

•

(6,0)

(0,5)

(2,3)

Objective Function: C = 5 x + 6 y

x ≥ 0

Contraints: y ≥ 0

x + y ≥ 5

3x + 4 y ≥ 18{

Vertices: C = 5 x + 6 y (0,5) C = 5 (0) + 6 (5) = 30

(2,3) C = 5 (2) + 6 (3) = 28

(6,0) C = 5 (6) + 6 (0) = 30

minimum value

No maximum value

x + y = 5 3x + 4y = 18

x = 0 Unbounded Region

y = 0

Find Maximum and Minimum Values

Ex 2

3.4

(5 , 3 , ─ 4 )+ y

+ x

+ z

─ z

─ x

─ y

( x , y , z )

•

5

─ 4

3

( x , y , z ) is an ordered triple, where 3 axes, taken 2 at a time, determine 3 coordinate planes that divide into eight octants.

Graphing Linear Equations in 3 variables 3.5

(3 , ─ 4 , ─ 2 )

+ y

+ x

+ z

─ z

─ x

─ y

( x , y , z )•

3

─ 4

─ 2

Linear equation in three variables, ( x , y , z ), is an ordered equation of the form: A x + B y + C z = D

[Where A, B, C and D are constants]

The graph of a an ordered triple is a plane.

Graphing Linear Equations in 3 variables 3.5

Plot points (4 , ─ 6 , 3 ) and (─ 7 , 5 , ─ 2) + y

+ x

+ z

─ z

─ x

─ y

4

•(4 , ─ 6 , 3 )

─ 7 ─ 6

5

3

─ 2 •(─ 7 , 5 , ─ 2)

(0, 7 , 0)

(0, 0, 4)

•

•(─ 5, , 0 , 0)

•

• (0, 0 , ─ 5)

•(0, ─ 7 , 0)

•(9, 0 , 0)

+ y

─ y

+ x─ x

─ z

+ z

•

•

•

•

•

•

••

A

B C

D

EF

• ( 3, 4, 2 )

( x, y, z )A = ( , , )

B = ( , , )C = ( , , )D = ( , , )E = ( , , )F = ( , , )

+ y

─ y

+ x─ x

─ z

+ z

•

•

•

•

•

•

••

A

B C

D

EF

• ( 3, 4, 2 )

( x, y, z )A = ( , , )

B = ( , , )C = ( , , )D = ( , , )E = ( , , )F = ( , , )

+ y

─ y

+ x─ x

─ z

+ z

3.5

(0, 6 , 0)

(0, 0, 3)

(4, , 0 , 0)

•

•

+ y

─ y

+ x─ x

─ z

+ z

•

3.5

(0, 6 , 0)

(0, 0, 3)

(4, , 0 , 0)

•

•

+ y

─ y

+ x─ x

─ z

+ z

3 x + 2 y + 4 z = 12

X Y Z

0 0 3

0 6 0

4 0 0

•

NOTES: Page 41, Section 3.5

Graphing Linear Equations in three variables ( 3 dimensions )

3.5

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

4 z = 12 – 3 x – 2 y

4 z = 12 – 3 x – 2 y 4 4

z = 12 – 3 x – 2 y 4

f (x,y) = 12 – 3 x – 2 y 4

3 x + 2 y + 4 z = 12

X Y Z

0 0 3

0 6 0

4 0 0

Replace z with f (x,y)

Evaluate function when x = 1 and y = 3 z = f (1 , 3 ) = 12 – 3 x – 2 y 4

z = f (1 , 3 ) = 12 – 3 (1) – 2 ( 3) 4

z = f (1 , 3 ) = 12 – 3 – 6 = 3 4 4

Graph has a solution = (1 , 3 , 3 ) 4

Evaluating a function of 2 variables as a function of x & y

3.5

1. If 3 planes intersect at a single point, the system has 1 solution2. If 3 planes intersect in a line, the system has infinitely many solutions3. If 3 planes have no point of intersection, the system has no solution

Linear Combination Method:1. Rewrite the linear system from 3 variables to 2 variables2. Solve the new linear system for both of its variables3. Substitute values found in step 2 into the original equation and solve for the

remaining variable

If you obtain an identity, such as 0 = 0, then the system has infinitely many solutions

If you obtain an identity, such as 0 = 1, in any of the steps, then the system has no solution

Graphing Linear Equations in 3 variables 3.5

3 x + 2 y + 4 z = 112 x − y + 3 z = 45 x − 3 y + 5 z = − 1

{{

3 x + 2 y + 4 z = 112 (2 x − y + 3 z = 4)

→ →

− 3 ( 2 x − y + 3 z = 4 ) 5 x − 3 y + 5 z = − 1

→ →

3 x + 2 y + 4 z = 114 x − 2 y + 6 z = 8

− 6 x + 3 y − 9 z = − 12 5 x − 3 y + 5 z = − 1

7 x + 10 z = 19

− x − 4 z = − 13

7 x + 10 z = 19 7 ( − x − 4 z = − 13)

→ →

7x + 10 z = 19− 7x − 28 z = − 91

− 18 z = − 72z = 4

7x + 10 z = 19

7x + 10 (4) = 19 7x + 40 = 19 7x = − 21 x = − 3

2 x − y + 3 z = 4

y = 2

( x , y , z )(− 3 , 2 , 4 )

2 (− 3) − y + 3 (4) = 4− 6 − y + 12 = 4

− y + 6 = 4− y = − 2

Solving Systems Using Linear Combination ( 1 solution )

3.6