"For HELPCLICK only!!"_BIO II and College Math_forum-B

profileAlphaMann
collegemath_1_7-for-helpclick.ppt

Section 1.7

Solving Inequalities

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Write each inequality using interval notation:

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Write each interval as an inequality involving x.

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Express as an inequality the result of multiplying each side of the inequality 3 < 5 by 2.

3(2) < 5(2)

6 < 10

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Note to keep this inequality true, the inequality symbol must be reversed.

Express as an inequality the result of multiplying each side of the inequality 3 < 5 by –2.

3(–2) ? 5(–2)

–6 > –10

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

]

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

)

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

)

(

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

]

[

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

(

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

In electricity, Ohm’s law states that E = IR, where E is the voltage (in volts). I is the current (in amperes), and R is the resistance (in ohms). An air-conditioning unit is rated at a resistance of 10 ohms. If the voltage varies from 110 to 120 volts, inclusive, what corresponding range of current will the air conditioner draw?

The air conditioner will draw between 11 and 12 amperes of current, inclusive.

Copyright © 2013 Pearson Education, Inc. All rights reserved

Copyright © 2013 Pearson Education, Inc. All rights reserved

)24

ax

-££

)27

bx

<<

)6

cx

³

)3

dx

<-

[

]

2,4

-

(

)

2,7

-

[6,)

¥

(,3]

-¥-

12

x

-<£

20

x

-££

5

x

>

1

x

<

)(1,2]

a

-

)(5,)

c

¥

)[2,0]

b

-

)(,1)

d

Solve the inequality: 531 and graph the

solution set.

x

-³-

5

531

5

x

-³-

--

36

x

-³-

36

33

x

£

--

--

2

x

£

{

}

2 or (,2]

xx

£

Solve the inequality: 4321 and graph th

e solution set.

xx

+<-

42

3

31

3

xx

<

-

--

+

{

}

2 or (,2)

xx

<--¥-

424

xx

<-

424

22

xx

xx

-

<-

-

24

x

<-

24

22

x

-

<

2

x

<-

Solve the inequality: 1325 and graph th

e solution set.

x

-<+<

132 and 325

xx

-<++<

{

}

11 or (1,1)

xx

-<<-

132

22

x

-

-+

-

<

33

x

-<

33

33

x

-

<

1

x

-<

2

3

2

25

x

+

-

<

-

33

x

<

33

33

x

<

and 1

x

<

52

13

3

Solve the inequality: and graph the so

lution set.

x

-

££

(

)

(

)

5

13

333

2

3

x

-

æö

££

ç÷

èø

{

}

21 or [2,1]

xx

-££-

3529

x

£-£

35

5

2

5

9

5

x

£-£

---

224

x

-£-£

22

22

4

2

x

-

-

-

³³

--

12

x

³³-

21

x

-££

(

)

1

Solve the inequality: 36>0 and graph th

e solution set.

x

-

+

(

)

1

1

36=0

36

x

x

-

+>

+

{

}

2 or (2,)

xx

>--¥

By the reciprocal property: 360

x

+>

36

x

>-

2

x

>-

If 32, find and so that 32.

xabaxb

-<<<+<

(

)

(

)

33332

x

-<<

32

x

-<<

936

x

-<<

923262

x

-+<+<+

7328

x

-<+<

So 7 and 8

ab

=-=

110120

E

££

110120

IR

££

110(10)120

I

££

110(10)120

101010

I

££

1112

I

££

