review 2.1-2.3. ex: check whether the ordered pairs are solutions of the system. x-3y= -5 -2x+3y=10...

Review 2.1-2.3Review 2.1-2.3

Ex: Check whether the ordered pairs Ex: Check whether the ordered pairs are solutions of the system.are solutions of the system.

x-3y= -5x-3y= -5-2x+3y=10-2x+3y=10

A.A. (1,4)(1,4)

1-3(4)= -51-3(4)= -5

1-12= -51-12= -5

-11 = -5-11 = -5

*doesn*doesn’’t work in the 1t work in the 1stst eqn, no need to check eqn, no need to check the 2the 2ndnd..

Not a solution.Not a solution.

B.B. (-5,0)(-5,0)

-5-3(0)= -5-5-3(0)= -5

-5 = -5-5 = -5

-2(-5)+3(0)=10-2(-5)+3(0)=10

10=1010=10

SolutionSolution

Solving a System GraphicallySolving a System Graphically

1.1. Graph each equation on the same Graph each equation on the same coordinate plane. (USE GRAPH PAPER!!!)coordinate plane. (USE GRAPH PAPER!!!)

2.2. If the lines intersect: The point (ordered If the lines intersect: The point (ordered pair) where the lines intersect is the pair) where the lines intersect is the solution.solution.

3.3. If the lines do not intersect:If the lines do not intersect:a.a. They are the same line – infinitely many They are the same line – infinitely many

solutions (they have every point in common).solutions (they have every point in common).

b.b. They are parallel lines – no solution (they They are parallel lines – no solution (they share no common points).share no common points).

Ex: Solve the system graphically.Ex: Solve the system graphically.2x-2y= -82x-2y= -82x+2y=42x+2y=4

(-1,3)

Ex: Solve the system graphically.Ex: Solve the system graphically.2x+4y=122x+4y=12

x+2y=6x+2y=611stst eqn eqn::

y = -½x + 3y = -½x + 3

22NDND eqn eqn::

y = -½x + 3y = -½x + 3

What does this mean?What does this mean?

the 2 equations are the 2 equations are for the same line!for the same line!Infinite Infinite many many solutionssolutions

ExEx: Solve graphically: x-y=5: Solve graphically: x-y=5 2x-2y=9 2x-2y=911stst eqn eqn::

y = x – 5y = x – 5

22ndnd eqn eqn::

y = x – 9/2y = x – 9/2

What do you notice about What do you notice about the lines?the lines?

They are parallel! They are parallel!

No solution!No solution!

Solving Systems of EquationsSolving Systems of Equations using using SubstitutionSubstitution

Steps:

1. Solve one equation for one variable

(y= ; x= ; a=)

2. Substitute the expression from step one into the other equation. Then solve.

3. Substitute back into Step 1 and solve for the other variable.

4. Check the solution in both equations of the system.

1) Solve the system using substitution1) Solve the system using substitutionx + y = 5x + y = 5

y = 3 + xy = 3 + x

Step 1: Solve an equation for one variable.

Step 2: Substitute

The second equation is

already solved for y!

x + y = 5x + (3 + x) = 5

2x + 3 = 5

2x = 2

x = 1

1) Solve the system using substitution1) Solve the system using substitutionx + y = 5x + y = 5

y = 3 + xy = 3 + x

Step 3: Plug back in to find the other variable.

y = 3 + x

Y = 3 + (1)

y = 4

Step 4: Check your solution.

(1, 4)

(1) + (4) = 5

(4) = 3 + (1)

The solution is (1, 4). What do you think the answer would be if you graphed the two equations?

Which answer checks correctly?Which answer checks correctly?

3x – y = 4x = 4y - 17

1.1. (2, 2)(2, 2)

2.2. (5, 3)(5, 3)

3.3. (3, 5)(3, 5)

4.4. (3, -5)(3, -5)

2) Solve the system using substitution2) Solve the system using substitution3y + x = 73y + x = 7

4x – 2y = 04x – 2y = 0


Step 2: Substitute

It is easiest to solve the

first equation for x.

3y + x = 7

-3y -3y

x = -3y + 7

4x – 2y = 0

4(-3y + 7) – 2y = 0


4x – 2y = 04x – 2y = 0


x = -3y + 7

x = -3(2) + 7

x = -6 + 7

x = 1

-12y + 28 – 2y = 0-14y + 28 = 0

-14y = -28y = 2


4x – 2y = 04x – 2y = 0


(1, 2)

3(2) + (1) = 7

4(1) – 2(2) = 0

When is solving systems by substitution easier to do than graphing?

When only one of the equations has a variable already isolated (like in example #1).

If you solved the first equation for x, what would If you solved the first equation for x, what would be substituted into the bottom equation.be substituted into the bottom equation.

2x + 4y = 43x + 2y = 22

1.1. -4y + 4-4y + 4

2.2. -2y + 2-2y + 2

3.3. -2x + 4-2x + 4

4.4. -2y+ 22-2y+ 22

3) Solve the system using substitution3) Solve the system using substitutionx = 3 – y x = 3 – y

x + y = 7x + y = 7


Step 2: Substitute

The first equation is

already solved for x!

x + y = 7

(3 – y) + y = 7

3 = 7

The variables were eliminated!!

This is a special case.

Does 3 = 7? FALSE!

When the result is FALSE, the answer is NO SOLUTIONS.

3) Solve the system using substitution3) Solve the system using substitution2x + y = 4 2x + y = 4

4x + 2y = 84x + 2y = 8


Step 2: Substitute

The first equation is

easiest to solved for y!

y = -2x + 44x + 2y = 8

4x + 2(-2x + 4) = 84x – 4x + 8 = 8

8 = 8This is also a special case.

Does 8 = 8? TRUE!

When the result is TRUE, the answer is INFINITELY MANY SOLUTIONS.

What does it mean if the result is What does it mean if the result is ““TRUETRUE””??

1.1. The lines intersectThe lines intersect

2.2. The lines are parallelThe lines are parallel

3.3. The lines are coincidingThe lines are coinciding

4.4. The lines reciprocateThe lines reciprocate

5.5. I can spell my nameI can spell my name

Solving Systems of EquationsSolving Systems of Equations using using EliminationElimination

Steps:

1. Place both equations in Standard Form

Ax + By = C.

2. Determine which variable to eliminate with Addition or Subtraction.

3. Solve for the variable left.

4. Go back and use the found variable in step 3 to find second variable.

5. Check the solution!!!!

1) Solve the system using elimination.1) Solve the system using elimination.

2x + 2y = 62x + 2y = 6

3x – y = 53x – y = 5Step 1: Put the equations in

Standard Form.

Step 2: Determine which variable to eliminate.

They already are!

None of the coefficients are the same!

Find the least common multiple of each variable.

LCM = 6x, LCM = 2y

Which is easier to obtain?

2y(you only have to multiplythe bottom equation by 2)



2(2) + 2y = 6

4 + 2y = 6

2y = 2

y = 1

2x + 2y = 6

3x – y = 5

Step 3: Multiply the equations and solve.

Multiply the bottom equation by 2

2x + 2y = 6

(2)(3x – y = 5)

8x = 16

x = 2

2x + 2y = 6(+) 6x – 2y = 10



(2, 1)

2(2) + 2(1) = 6

3(2) - (1) = 5

2x + 2y = 6

3x – y = 5

Solving with multiplication adds one more step to the elimination process.


x + 4y = 7x + 4y = 7

4x – 3y = 94x – 3y = 9Step 1: Put the equations in

Standard Form.They already are!



LCM = 4x, LCM = 12y


4x(you only have to multiplythe top equation by -4 to

make them inverses)


x + 4y = 7x + 4y = 7

4x – 3y = 94x – 3y = 9


x + 4(1) = 7

x + 4 = 7

x = 3


Multiply the top equation by -4

(-4)(x + 4y = 7)

4x – 3y = 9)

y = 1

-4x – 16y = -28 (+) 4x – 3y = 9

-19y = -19



(3, 1)

(3) + 4(1) = 7

4(3) - 3(1) = 9

x + 4y = 7

4x – 3y = 9

What is the first step when solving with What is the first step when solving with elimination?elimination?

1.1. Add or subtract the equations.Add or subtract the equations.

2.2. Multiply the equations.Multiply the equations.

3.3. Plug numbers into the equation.Plug numbers into the equation.

4.4. Solve for a variable.Solve for a variable.

5.5. Check your answer.Check your answer.

6.6. Determine which variable to Determine which variable to eliminate.eliminate.

7.7. Put the equations in standard form.Put the equations in standard form.

Which variable is easier to eliminate?Which variable is easier to eliminate?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

3x + y = 44x + 4y = 6

1.1. xx

2.2. yy

3.3. 66

4.4. 44


3x + 4y = -13x + 4y = -1

4x – 3y = 74x – 3y = 7

Step 1: Put the equations in Standard Form.

They already are!



LCM = 12x, LCM = 12y


Either! I’ll pick y because the signs are already opposite.


3x + 4y = -13x + 4y = -1

4x – 3y = 74x – 3y = 7


3(1) + 4y = -1

3 + 4y = -1

4y = -4

y = -1


Multiply both equations

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

(4)(4x – 3y = 7)

x = 1

9x + 12y = -3 (+) 16x – 12y = 28

25x = 25



(1, -1)

3(1) + 4(-1) = -1

4(1) - 3(-1) = 7

3x + 4y = -1

4x – 3y = 7

What is the best number to multiply the top What is the best number to multiply the top equation by to eliminate the xequation by to eliminate the x’’s?s?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

21 22 23 24 25 26 27 28 29 30 31 32

3x + y = 46x + 4y = 6

1.1. -4-4

2.2. -2-2

3.3. 22

4.4. 44

Solve using elimination.Solve using elimination.

2x – 3y = 1x + 2y = -3

1.1. (2, 1)(2, 1)

2.2. (1, -2)(1, -2)

3.3. (5, 3)(5, 3)

4.4. (-1, -1)(-1, -1)

Find two numbers whose sum is 18 Find two numbers whose sum is 18 and whose difference 22. and whose difference 22.

1.1. 14 and 414 and 4

2.2. 20 and -220 and -2

3.3. 24 and -624 and -6

4.4. 30 and 830 and 8

Using Elimination to Solve a Using Elimination to Solve a Word Problem:Word Problem:

Two angles are supplementary. The measure of one angle is 10 degrees more than three times the other. Find the measure of each angle.


Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle.

x = degree measure of angle #1

y = degree measure of angle #2

Therefore x + y = 180


Two angles are supplementary. The measure of one angle is 10 more than three times the other. Find the measure of each angle.

x + y = 180x =10 + 3y


Solvex + y = 180x =10 + 3y

x + y = 180

-(x - 3y = 10)

4y =170

y = 42.5

x + 42.5 = 180 x = 180 - 42.5

x = 137.5

(137.5, 42.5)


The sum of two numbers is 70 and their difference is 24. Find the two numbers.

Using Elimination to Solve a Using Elimination to Solve a Word problem:Word problem:


x = first number

y = second number

Therefore, x + y = 70



x + y = 70

x – y = 24


x + y =70

x - y = 24 2x = 94

x = 47

47 + y = 70

y = 70 – 47

y = 23

(47, 23)

Now you Try to Solve These Now you Try to Solve These Problems Using Problems Using Elimination.Elimination.

Solve1. Find two numbers whose sum is

18 and whose difference is 22.

2. The sum of two numbers is 128 and their difference is 114. Find the numbers.

MATRIX:MATRIX: A rectangular A rectangular arrangement of arrangement of numbers in rows and numbers in rows and columns.columns.

The The ORDERORDER of a matrix of a matrix is the number of the is the number of the rows and columns.rows and columns.

The The ENTRIESENTRIES are the are the numbers in the matrix.numbers in the matrix.

502

126rows

columns

This order of this matrix This order of this matrix is a 2 x 3.is a 2 x 3.

67237

89511

36402

3410

200

318 0759

20

11

6

0

7

9

3 x 3

3 x 5

2 x 2 4 x 1

1 x 4

(or square matrix)

(Also called a row matrix)

(or square matrix)

(Also called a column matrix)

To add two matrices, they must have the same To add two matrices, they must have the same order. To add, you simply add corresponding order. To add, you simply add corresponding entries.entries.

34

03

12

70

43

35

)3(740

0433

13)2(5

44

40

23

9245

3108

2335

2571

)1(8 70 51 23

55 34 32 )2(9 =

= 7 7 4 5

0 7 5 7

To subtract two matrices, they must have the same To subtract two matrices, they must have the same order. You simply subtract corresponding entries.order. You simply subtract corresponding entries.

232

451

704

831

605

429

2833)2(1

)4(65015

740249

603

1054

325

724

113

810

051

708

342

=

5-2

-4-1 3-8

8-3 0-(-1) -7-1

1-(-4)

2-0

0-7

=

2 -5 -5

5 1 -8

5 3 -7

In matrix algebra, a real number is often called a In matrix algebra, a real number is often called a SCALARSCALAR. . To multiply a matrix by a scalar, you multiply each entry in To multiply a matrix by a scalar, you multiply each entry in the matrix by that scalar. the matrix by that scalar.

14

024

416

08

)1(4)4(4

)0(4)2(4

86

54

30

212

)8(360

52412

-2

6

-3 3

-2(-3)

-5

-2(6) -2(-5)

-2(3) 6 -6

-12 10

Multiplying MatricesMultiplying Matrices

In order to multiply matrices...In order to multiply matrices...

A • B = ABm x n n x p m x

p

Can you multiply? What will the dimensions be?

2 x 3 3 x 4

AB

5 x 3 5 x 2

2 x 4

Not possible

Ex 1.

A B

A B AB

How to multiply...How to multiply...

b

a c d

ac=

ad

2 x 1 1 x 2 2 x 2


b

a c d

ac=

ad

2 x 1 1 x 2 2 x 2

bc

bd

Ex. 1Ex. 1

5

2 1 3

Ex. 2Ex. 2

20

14A

21

34B

-16 + 1

0 - 4

-12+2

0 - 2

-15

- 4-2

-10=

Find AB

Ex. 3Ex. 3

33

12A

23

10B

Find BA

23

10

30

33

33

12


fed

cba

mk

ji

hg ag=

+bi +ck ah

+bj +cm

2 x 3 3 x 2 2 x 2


fed

cba

mk

ji

hg ag=

+bi +ck

dg

+ei +fk

ah

+bj +cm

dh

+ej +fm

2 x 3 3 x 2 2 x 2

Ex. 4Ex. 4

40

25

51

86

34

-1(4)+5(6)

5(-3) +2(8)5(4)+2(6)

0(4)+-4(6) 0(-3)+-4(8)

-1(-3) +5(8) -4+ 30

-15+ 1620 +12

0 + -24 0 + -32

3 + 40

26

132

-24 -32

43

=

If we are multiplying matrices, we multiply each row of the first matrix by each column in the second matrix!!

multiply each row of the first matrix by each column in the second matrix!!

3 1 11

1 2 22

1 0 51

4 1 2

1

2

1

3*1 1*2 1*1

1*1 2*2 2*1

1*1 0*2 5*1

4*1 1*2 2*1

+

+

+

+

+

+

+

+

3 9 12 3 5

2.) 8 0 41 6 8

3 1 5

2 X 3

3 X 3

2(3)+(-3)(8)+5(-3)

2(9)+(-3)(0)+ 5(1) 2(1)+(-3)(-4)+ 5(5)

-1(3)+(6)(8)+8(-3) -1(9)+(6)(0)+ 8(1) -1(1)+(6)(-4)+ 8(5)

33 23 39

21 1 15

3 9 11 0 4

3.) 8 0 42 5 0

3 1 5

9 13 21

46 18 18

Ex. 5Ex. 5

Examples:

2(3) + -1(5) 2(-9) + -1(7) 2(2) + -1(-6)

3(3) + 4(5) 3(-9) + 4(7) 3(2) + 4(-6)

3 9 2 2 1.

5 7 6 3 4

2

Dimensions: 2 x 3 2 x 2

*They don’t match so can’t be multiplied together.*

1 2 1 1

1 3 2 7

2 6 1 8

x

y

z

3.

0 1 4 3

1 0 2 5

4.

2 x 2 2 x 2

*Answer should be a 2 x 2

0(4) + (-1)(-2) 0(-3) + (-1)(5)

1(4) + 0(-2) 1(-3) +0(5)

Solving Systems of EquationsSolving Systems of Equationswith Matriceswith Matrices

A system of equations may be represented as a matrix equation. For example, the system of equations

22

153

yx

yx

may be represented by the matrix equation

2

1

12

53

y

x

Write the matrix equation that represents the system:

12

8

yx

yx

1

8

12

11

y

x

Write the matrix equation that represents the system:

92

133

yx

yx

9

13

12

31

y

x

A matrix equation is in the form AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

2

1

12

53

y

x

12

53A

y

xX

2

1B

Real Numbersax=b(1/a)(ax) = (1/a)b(๋1/a)(a)x = b/a1x = b/ax = b/a

Solving AX=BSolving AX=B

Note: 1/a must exist to solve ax = b

Real Numbersax=b(1/a)(ax) = (1/a)b(๋1/a)(a)x = b/a1x = b/ax = b/a

MatricesAX=BA-1(AX)=A-1B(A-1A)X=A-1BIX=A-1BX=A-1B

Solving AX=BSolving AX=B

Note: A-1 must exist to solve AX=B

Solve the system of equations using matrices.

12

8

yx

yx

x = -7

y = 15

Solve the system of equations using matrices.

x = 1

y = -1

Ex. 2Ex. 2 Solve using matrices. Solve using matrices.

3 4 5

2 10

x y

x y

x = -7

y = -4

10

5

12

43

y

x

A B

X = AA-1-1B

AX = B

(-7, -4)

Ex. 3Ex. 3 Solve using matrices Solve using matrices

7 3 11

14 4 2

x y

x y

x = 5/7

y = 2

(5/7, 2)

review 2.1-2.3. ex: check whether the ordered pairs are solutions of the system. x-3y= -5 -2x+3y=10...

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