mba10_ppt_0203.ppt

21
Chapter Chapter 2 2 Section Section 3 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Transcript of mba10_ppt_0203.ppt

Page 1: mba10_ppt_0203.ppt

Chapter Chapter 22Section Section 33

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Page 2: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

More on Solving Linear Equations

Learn and use the four steps for solving a linear equation.Solve equations with fractions or decimals as coefficients.Solve equations with no solution or infinitely many solutions.Write expressions for two related unknown quantities.

11

44

33

22

2.32.32.32.3

Page 3: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Objective 11

Slide 2.3 - 3

Learn and use the four steps for solving a linear equation.

Page 4: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Learn and use the four steps for solving a linear equation.

Slide 2.3 - 4

We solve more complicated equations using the following four-step method.

Step 1: Simplify each side separately. Clear parentheses, using the distributive property if needed, and

combine all like terms.

Step 3: Isolate the variable. Use the multiplication property if necessary to get the equation in the form x = a number. (Other letters may be used for variables.)

Step 2: Isolate the variable term on one side. Use the addition property if necessary so that the variable term is on one side of the equation and a number is on the other.

Step 4: Check. Substitute the proposed solution into the original equation to see if a true statement results.

Page 5: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 1

Slide 2.3 - 5

Solve.

Solution:

Using the Four Steps to Solve an Equation

510 5 55x

10

1 0

10

0 1x

1x

Check:

3 5 6 11 2 5x x x

8 5 2 5x x 28 5 2 5 2x xx x

3 5 6 11 2 5x x x 3 5 6 11 2 11 5

3 5 6 11 2 5 3 3

The solution set of the equation is {1}.

Page 6: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2

Slide 2.3 - 6

Solve.

Solution:

Using the Four Steps to Solve an Equation

1414 1 142 6a

2

2 2

2

a

1x

Check:

11 3 1 5 16a a

11 3 3 5 16a a 51 3 1 54 5 6a aa a

11 3 1 5 16a a

11 3 1 51 161

11 3 0 5 16

11 11

The solution set of the equation is {−1}.

Page 7: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 3

Slide 2.3 - 7

Solve.

Solution:

Using the Four Steps to Solve an Equation

73 77 9x

3 6

3 3

1x

16

3x

Check:

4 ( 7) 9x x

7) 914 (x x

4 7 9x x

4 ( 7) 9x x

4 716

39

16

3

64 16 21 27

3 3 3 3 64 37 27

3 3 3

The solution set of the equation is16

.3

27 27

3 3

Page 8: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 3

Slide 2.3 - 8

Solve.

Using the Four Steps to Solve an Equation

4 ( 7) 9x x

Be very careful with signs when solving an equation like the one in the previous example. When clearing parentheses in the expression remember that the − sign acts like a factor of −1 and affects the sign of every term within the parentheses.

Page 9: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4

Slide 2.3 - 9

Solve.

Solution:

Using the Four Steps to Solve an Equation

4 1 422 4 8z

22 2

2

2

22 2z

1z

Check:

2 3 2 6 4 1 14z z

2 6 18 4 4 14z z

44 18 4 18 4z zz z 2 3 2 6 4 1 141 1 2 3(2 6) 4 0 14

2 3( 4) 14 2 12 14

The solution set of the equation is {−1}.

2 3 2 6 4 1 14z z

14 14

Page 10: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Objective 22

Slide 2.3 - 10

Solve equations with fractions or decimals as coefficients.

Page 11: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

We clear an equation of fractions by multiplying each side by the least common denominator (LCD) of all the fractions in the equation. It is a good idea to do this to avoid messy computations.

Solve equations with fractions or decimals as coefficients.

Slide 2.3 - 11

When clearing an equation of fractions, be sure to multiply every term on each side of the equation by the LCD.

Make sure to multiply each term of the equation by the LCD. It is best to identify how many terms are in the equation before multiplying.

Page 12: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 5

Slide 2.3 - 12

Solve.

Solution:

Solving an Equation with Fractions as Coefficients

5 99 92x 14 2

2 2

x

7x

Check: 1 5 3 1

3 12 4 2x x

44 5 6 49xx x x

33 33

12 12

The solution set of the equation is {−7}.

1 5 3 1

3 1212 12 12

21

42x x

1 5 3 1

3 12 27 7

4

7 5 3 7

3 12 4 2

28 5 9 42

12 12 12 12

1 5 3 1

3 12 4 2x x

Page 13: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6

Slide 2.3 - 13

Solve.

Solution:

Solving an Equation with Decimals as Coefficients

10 15 38x xx x

14 1

2

4

14 8x

2x

0.5 2 3 4.5 0.1 7x x

0.5 2 3 4.5 0.1 710 10 10x x

5 2 5 3 45 7x x

1010 13 014 8x

Page 14: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 6

Slide 2.3 - 14

Solving an Equation with Decimals as Coefficients (Check)

Check: 0.5 2 3 4.5 0.1 7x x

10 5 6 47 7

10 30 40 40 40

The solution set of the equation is {−2}.

5 2 5 3 45 72 2

Page 15: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Objective 33

Slide 2.3 - 15

Solve equations with no solution or infinitely many solutions.

Page 16: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Each equation that we have solved so far has had exactly one solution. An equation with exactly one solution is a conditional equation because it is only true under certain conditions. Sometimes equations may have no solution or infinitely many solutions.

Solve equations with no solution or infinitely many solutions.

Slide 2.3 - 16

DO NOT write { 0 } as the solution set. There are infinitely many other solutions.

Page 17: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 7

Slide 2.3 - 17

Solve.

Solution:

Solving an Equation That Has Infinitely Many Solutions

10 10 01 10 0 0

3 10 2 4 14x x x

2 10 2 10x x 22 10 1 22 0xx x x

The variable has “disappeared.” Since the last statement is true, any real number is a solution. An equation with both sides exactly the same, is called an identity. An identity is true for all replacements of the variables. As shown above, write the solution set as {all real numbers}.

Page 18: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 8

Slide 2.3 - 18

Solve.

Solution:

Solving an Equation That Has No Solution.

8 6

3 8 6 1 3x x x

3 8 6 6 3x x x 33 38 3 6xx xx

Again, the variable has “disappeared,” but this time a false statement results. When this happens in solving an equation, it indicates that the equation has no solution and is called a contradiction. Its solution set is the empty set, or null set, symbolized Ø.

DO NOT write { Ø } to represent the empty set.

Page 19: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The following table summarizes the solution sets of the three types of equations presented so far.

Solve equations with no solution or infinitely many solutions.

Slide 2.3 - 19

Page 20: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Objective 44

Slide 2.3 - 20

Write expressions for two related unknown quantities.

Page 21: mba10_ppt_0203.ppt

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 9

Slide 2.3 - 21

Solution:

Translating a Phrase into an Algebraic Expression

One number is 5 more than twice another. If the first number is represented by x, write an expression for the second number.

2 5x