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1 1 1 1 2 2 2 2 3 3 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3 3 3 3 3 2 2 2 1 1 1

1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

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Page 1: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

11 11

22 22

33

Solve equations in 2 variable

Solve equations in 3 variables

Matrices Operations

Word problems

3 33 33

222

111

Page 2: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Solving the first equation for x, we find x = 3y – 1. Substituting this value of x into the second equation yields

Substituting this value of y into the first equation gives . Therefore, the unique solution of the system is (2, 1).

Solve the system of equations:

3 1

4 3 11

x y

x y

4 3 1 3 11

12 4 3 11

1

( )

.

y y

y y

y

or

Page 3: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Solve the system of equations

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 3

2 3 8

x y z

x y z

x y z

3 2 8 9

2 2 1 3

1 2 3 8

3 2 8 9

2 2 1 3

1 2 3 8

1 0 0 3

0 1 0 4

0 0 1 1

1 0 0 3

0 1 0 4

0 0 1 1

The solution to the system is thus x = 3, y = 4, and z = 1.

Page 4: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

0 2 4

2 3 7

3 1 8

A

Given the matrix find 5A.

5(0) 5(2) 5(4)

5 5( 2) 5(3) 5(7)

5(3) 5(1) 5(8)

A

0 10 20

10 15 35

15 5 40

Page 5: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

The market share of motorcycles in the United States in 2001 follows: Honda 27.9%, Harley-Davidson 21.9%, Yamaha 19.2%, Suzuki 11%, Kawasaki 9.1%, and others 10.9%. The corresponding figures for 2002 are 27.6%, 23.3%, 18.2%, 10.5%, 8.8%, and 11.6%, respectively. Express this information using a 2x6 matrix.

H H-D Y S K O

2001 27.9 21.9 19.2 11 9.1 10.9

2002 27.6 23.3 18.2 10.5 8.8 11.6A

Page 6: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Solving the first equation for x, we obtain

Substituting this value of x into the second equation, we obtain

Since this is impossible, we conclude that the system of equations has no solution.

Solve the equation

3 7 473

43

x y

x y

9 12 1473

43( ) y y

21 12 12 14y y

21 14.

3 4 7

9 12 14

x y

x y

Page 7: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Solve the system 3 1

3 2 4 3

x y z

x y z

1 1 3 1 1 0 2 1

3 2 4 3 0 1 1 0

If we let z = t then the solution is given by

(1 + 2t, t, t)

Page 8: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Given matrices A and B, find A + 2B0 2 4

2 3 7

3 1 8

A

1 3 0

3 6 7

2 9 1

B

2 8 4

2 8 15 21

7 19 6

A B

Page 9: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

We wish to solve the system of equations x + y = 2000 (x = the amount invested at 6 percent) 0.06x + 0.08y = 144 (y = the amount invested at 8 percent)

Using the Gauss-Jordan method, we find

the solution to this system of equations is x = 800 and y = 1200.

Michael Perez has a total of $2000 on deposit with two savings institutions. One pays interest at the rate of 6%/year, whereas the other pays interest at the rate of 8%/year. If Michael earned a total of $144 interest during a single year, how much does he have on deposit in each institution?

2 1 20.06 501 1 2000 1 1 2000 1 1 2000

0.06 0.08 144 0 0.02 24 0 1 1200R R R

1 21 0 800

0 1 1200R R

Page 10: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Solving the first equation for x, we have5x = 6y + 8

Substituting this value of x into the second equation, gives 12y + 16 – 12y = 16or 0 = 0.Thus, by assigning the value t to x, where t is any real number, we find that

. So the ordered pair, is a solution to the system.

Solve the equation

y t 56

43

( , )t t56

43

5 6 8

10 12 16

x y

x y

Page 11: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Solve the system2 3

2 3 2

2 2 4 5

x y z

x y z

x y z

1 1 2 3 1 0 1 11/ 5

2 3 1 2 0 1 1 4 / 5

2 2 4 5 0 0 0 1

Notice the false statement 0 = -1

The system is inconsistent and has NO solution.

Page 12: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

Solve for u, x, y and z in the matrix2 2 3 2

2 4 2

2 3 2

3 2

2 4 5

4 3 2

x

y

z

u

L

NMMM

O

QPPP

L

NMMM

O

QPPP

by the definition of equality of matrices,

u

x x x

y y

z z

3

2 2 3 2 5 5 2

2 5 7

2 4 2

and or

and

and

, / ,

, ,

, .

Page 13: 1111 1111 2222 2222 3333 Solve equations in 2 variable Solve equations in 3 variables Matrices Operations Word problems 3333 3333 3333 2222 2222 1111

We see that x = 80,000, y = 20,000, and z = 100,000. Therefore, they should invest $80,000 in high-risk, $20,000 in medium-risk, and $100,000 in low-risk stocks.

A private investment club has $200,000 earmarked for investment stocks. To arrive at an acceptable overall lever of risk, the stocks that management is considering have been classified into three categories: high risk, medium risk, and low risk. Management estimates that high-risk stocks will have a rate of return of 15%/year; medium-risk stocks, 10%/year; and low-risk stocks, 6%/year. The members have decided that the investment in low-risk stocks should be equal to the sum of the investments in the stocks of the other two categories. Determine how much the club should invest in each type of stock if the investment goal is to have a return of $20,000/year on the total investment.

200,000

0.15 0.1 0.06 20,000

0

x y z

x y z

x y z

Let x, y, and z denote the amount to be invested in high-risk, medium-risk, and low-risk stocks, respectively.