ch09 math textbook

Fractions and Fractional Equations

Prepared by: Richard Mitchell Humber College

9

http://en.wikipedia.org/wiki/Algebraic_fraction

CASE STUDY

9.1-SIMPLIFICATION OF FRACTIONS

9.1-DEFINITIONS-Pages 227 to 231

C o n s t a n t s a n d V a r i a b l e s

P ro p er an d Im p ro p er F rac tio n s

See WileyPLUS glossary for

terms and definitions Mixed FractionsAlgebraic Fractions

N u m e r a t o r s

D e n o m i n a t o r s

T e r m sL o w e s t

F r a c t i o n s a n d D e c i m a l s

Common Fractions

9.1-EXAMPLE 12(b)-Page 230Reduce to lowest terms. Give the answer with positive exponents.

2ANS:

3

x

yz

2

2 3

3

9

x yz

xy z

2 1 1

1 2 3

3

9

x y z

x y z

1

1 2

1

3

x

y z

9.1-EXAMPLE 13(a)-Page 230Reduce to lowest terms. Give the answer with positive exponents.

2 1ANS:

3

x

22

3

x x

x

(2 1)

3

xx

x

9.1-EXAMPLE 13(c)-Page 230Reduce to lowest terms. Give the answer with positive exponents.

3ANS:

2 1

x

x

2

2

2 5 3

4 1

x x

x

( ) ( 3)

( )

2 1

1 (2 1)2

x

x

x

x

9.1 EXAMPLE 13(d)-Page 230Reduce to lowest terms. Give the answer with positive exponents.

2ANS:

2 3

x b

x a

2

2 2

( ) (2 2 )

2 3

x ax bx ab

x ax a

)32()(

)(2)(

axax

axbaxx

( ) ( 2 )

( ) (2 3 )

x a

x a

x b

x a

9.1-EXAMPLE 14-Page 231Simplify the fraction.

A N S : 1

3 2

2 3

x

x

3 2

2 3

x

x

3 2

2 3

x

x

3 2

3 2

x

x

1

1

9.2-MULTIPLICATION AND DIVISION OF FRACTIONS

9.2-EXAMPLE 16(b)-Page 233Multiply and reduce to lowest terms.

2

2ANS:

x -

x

2

3

4

2

x x

x x

1 2

3

( ) ( 4)

( ) ( 2)

x x

x x

1

3

( ) ( ) ( 2)

( ) (

2

2)

x x

xx

x

1

3

( ) ( 2)

( )

xx

x

9.2-EXAMPLE 16(c)-Page 233Multiply and reduce to lowest terms.

A N S : 1

2 2

2 2

2 2 3 9

4 3 2 7 6

x x x x

x x x x

)3()1(

)2()1(

xx

xx

)32()2(

)32()3(

xx

xx

( ) ( )

( ) ( )

1

3

2

1

x

x

x

x

( ) ( )

( ) (

2

2

3 3

32 )x

x x

x

9.2-EXAMPLE 17(d)-Page 234Divide and reduce to lowest terms.

A N S : ( 1 )x x -

2

2

2 2x x x

x x

2 22

2

x x x

x x

2

1

( ) ( 1)

(

2

2)

xx x

xx

9.2-EXAMPLE 17(e)-Page 234Divide and reduce to lowest terms.

2

4ANS:

r

2

4

r xx

2

4

1

x

xr

9.3-ADDITION AND SUBTRACTION OF FRACTIONS

9.3-EXAMPLE 25-Page 238Combine and simplify.

2 10ANS:

2

x y

xy

xy

x 5

2 2

22

x

x

x

y xy

x

L.C.D. = 2xy

2

05 2 1

2

y

y xx y

y

2 1

2 2

0

xy xy

x y

2 10

2x

x

y

y

Rough Work


32 5ANS:

2

x

x

32 2

2

2

1 2

x

x

x

x

x

L.C.D. = 2x5

2 2

51

1x x

32 5

2 2x x

x

3

2

2 5

x

x

Rough Work

2 5

2x

x


2

2

5 7ANS:

3 11 6

x x

x x

23 2

3 2

( 2) ( )

( 3) ( )(3 3

4

)

3 2 6

2

x

x x

x xx

xx

x

2(2 1) ( )3

3(

2

3 2 3) ( )( )3 2

6 3x

x

x

x x

x

x

xx

2 2( ) ( )

( )( ) ( )

3 2

3 3 2 3 ( )2

4 2

3

6 6 3x x x x x

x x

x

x x

Rough Work

23

12

3

2

x

x

x

x

L.C.D. = (x – 3)(3x – 2)

2 2(3 2 6 4) (2 6 3)

( )(3 3 2)

x x x x x

x

x

x

9.3-EXAMPLE 22-Page 237 (optional)

Combine and simplify.

2

( )( 1)

( )(

5 ( ) ( )

( )( 1)

5 1

( )( )( )1 11)

x x

x x

x x

xx x x x

2

22

2

( ) ( )

( 1)( 1) ( )( )( 1)1

x

x x x xx

x

x

xx

Rough Work

2 2 3 2

5 9

1

x

x x x x x

22

9 ( )

( )( 1) ( )( )(

( 1)

( 1 ))

9

1 1

1x

x xx x x

x

x

L.C.D. = (x2)(x + 1)(x – 1)

(con’t)

9.3-EXAMPLE 22-Page 237 (optional)

Combine and simplify.

2

35 ( 1) ( ) 9( 1)

( ( 1) )1

x x x x

x x x

2 3

2

5 5 9 9

)( )1 1(x x

x x x x

x

3 2

2

5 4 9ANS:

( 1)( 1)

x x x

x x x

2 2

2

2

( ) ( ) ( ) ( ) ( )

( )( )( ) (

5 1 9

)( )( ) ( )( )(1 1 1 1 1)1

1

x x x x x

x x x x x

x x x x

9.4 – COMPLEX FRACTIONS

9.4-EXAMPLE 33-Page 243 (long way)

Simplify the complex fraction.

14

ANS: 39

1 2

2 31

34

1 2

2 33 11 4

3 4

12 16 6

4 4

7

6

4

13

37

6

1

4

1

7

6

4

3

1

7

3

2

3

14

39

L.C.D. = 6 and 4

9.4-EXAMPLE 33-Page 243 (short way)


14

ANS: 39

1 2

2 31

34

12 1 2

32

112

3

1 4

6 8

36 3

14

39

L.C.D. = 12

9.4-EXAMPLE 34-Page 243 (long way)


( )

ANS: ( )a a bb a b

1

1

abba

11

1

1

abba

b bb a

aab

a

b aba ba

ab

a

a

b

b

ab

b

a

b a

( )

( )

b a

b

a

a b

L.C.D. = a and b

9.4-EXAMPLE 34-Page 243 (short way)


( )

ANS: ( )a a bb a b

1

1

abba

11

11

abba

bab

a

2

2

( )

( )

ab a a a bor

b a bab b

L.C.D. = ab

9.4-EXAMPLE extra


(2 1)

ANS: 2

a

2 1

412

a

a

2 1

4

1 2

11

a

a

2

4 44 1

2 2

2 1

a

a

24 1

42 1

2

a

a

24 1

4

2 1

2

a a

2 24 1

4

2

2 1

a

a

( )2

2 1

2 1 2( )

)4(

1a a

a

(2 1)

2

a

L.C.D. = 2 and 4

2

9.5 – FRACTIONAL EQUATIONS

9.5-STRATEGY-Pages 245 to 247 Eliminate Fractions:

Multiply both sides of the equation by the lowest common denominator.

Remove Parenthesis:Brackets are multiplied away.

Collect x Terms:Move all x terms to one side and all other terms to other side.

Combine Like Terms:Always simplify.

Remove Coefficient of x:Divide both sides by coefficient.

Check Answer:Be sure to substitute the answer back into the original equation.

9.5-EXAMPLE 36-Page 246Solve the equation for x:

9x – 5x = 2

4x = 2

x = 1/2

CHECK:

5 3 15

3 2x x

3 25 15

5 153

1x x

3 25 3 1

15 155

15x x

1 12 2

32

5 3 15

3 1 2

10 6 15

checks2 2

( )15 15

1ANS:

2

LCD = 15


4 = 30 + 3x

3x = -26

x = -26/3

CHECK:

2 5

2

1

3x x

6 62 5 1

3 2x x

x x

2 5 16

36 6

2x x x

x x

2 5 1

2326 263 3

2 15 13

26 26 26

checks2 2

( )26 26

26ANS: -

3

LCD = 6x


8 7 2

5 4 1

2

5 1

x x

x x

8 7 2 2( )( ) ( )( ) ( )( )5 4 5 1 5 4 5 1 5 4 5 1

5 4 1 5 1x x x x x x

x x

x x

( 8 7 ) ( 5 1 ) 2 ( 5 4 ) ( 5 1 ) 2 ( 5 4 )x x x x x x

LCD =(5x+4)(5x+1)

2 2 24 0 3 5 8 7 2 ( 2 5 5 2 0 4 ) 1 0 8x x x x x x x x

2 2 24 0 4 3 7 5 0 5 0 8 1 0 8x x x x x x

4 3 7 4 2 8x x

1x (con’t)


CHECK:

8 7 22

5 4 5 1

x x

x x

8( ) 7 2( )2

5( ) 4 5( ) 1

1 1

1 1

A N S : 1

15 22

9 6

(c )hecks5 5

3 3


23 12

3

3

2 4

x xx x

3 1 3 1 33 2 4

( )( ) ( )( ) ( )( )3 ( 3)( 1) 1

1x x

x x xx x

x x x

3 ( 1 ) 2 4 ( 3 )x x

LCD =(x-3)(x+1)

3 3 4 1 0x x

1 3x (con’t)

3 2 4( )(3 )3 1 1x x x x


CHECK:

2

3 2 4

3 12 3x xx x

2

3 2 4

( ) 3 ( ) 1( ) 2( ) 313 1313 13

A N S : 1 3

3 2 4

10 140 14

che( )cks3 3

10 10

9.6-WORD PROBLEMS LEADING TO FRACTIONAL EQUATIONS

9.6-STRATEGY-Pages 249 to 250 Rate Problems (Uniform Motion, Work, Fluid or Energy)

9.6-EXAMPLE 41-Page 249 (Uniform

Motion)A train departs at noon travelling at a speed of . A car leavesthe same station later to overtake the train, travelling on a roadparallel to the track. If the car's speed is

1/264 km/h

96 km/ h

h, at what time and at what distance from the station will it overtake the train ?

ANS: 1:30 p.m. 96 km

RATE TIME DISTANCE

Train

Car

TOTALS

x =

(t)

(t + ½)

-

64

96 96 (t)

64 (t + ½)

1

26 )4 9 (6 t t

64 32 96t t

hour (time for car)1 t

Equation

- Distance=Distance

hour

1 1

2 2

(time for train)1 t

=

9.6-EXAMPLE 42-Page 250 (Work)

Crew A can assemble , and Crew B can assemble . If both crews together assemble with Crew B working

longer

2 cars in 5

than Crew A

days 3 cars in7 day

, how many days mu1

s0

d t 100

eac cars,

h creway s

s work ?

days (Crew A)

days (Crew B)

ANS: 116 126

RATE TIME Amount of Work

Crew A

Crew B

TOTALS

x =

(t + 10)

(t)

-

2/5

3/7 3/7 (t + 10)

2/5 (t)

2 3

5 7( ) ( ) 10010t t

14 15 150 3500t t

days (Crew A)116 t

Equation

- 100

days (Crew B)10 126 t

=

9.6-EXAMPLE 43-Page 251 (Fluid or

Energy Flow)

4.0 mA sma

onthsll hydroelectric generating station can produce

After of operation, another generator isadded which, by it

61 gagajoules (GJ) of energy per year.

39 se GJlf, in can prod 5.0 montuce

hs. How many additional are needed 9months 5 GJfor a total of to be pro duced ?

monthsANS: 5.8

RATE TIME Amount of Flow

Gen 1

Gen 2

TOTALS

x =

(t)

(t + 4)

-

61/12

39/5 39/5 (t)

61/12 (t + 4)

( )61 39

1) 9

254

5(t t

305 1220 468 5700t t

months (Generator 2)5.8 t

Equation

- 95

(5.8 additional months needed to reach 95 GJ).

=

9.7-LITERAL EQUATIONS AND FORMULAS


CHECK:

b b dxa

db

x

a b

x db

a b

b b da

da bb

db b b d

b

che( )cks d

b db

ANS: d

x a bb

Divide both sides by b

Subtract b from both sides

Multiply both sides by a

da bxb

9.7-EXAMPLE 48-Page 255

CHECK:

1 2( )kA tq

t

L

12qL

t tkA

2 2qL

tk

qA

kA t

L

2 2qL kAt kAtq

L

c h e c k( s ) q q

1 2ANS: qL

t tkA

Multiply both sides by L/kA

Add t2

to both sides

The formula for the amount of heat flowing by conduction througha wall of thickness , conductivity , and cross-sectional area is

qL k A

(where and are the temperatures of 1 2the warmer and cooler sides, respectively)

1

.

Solve this equation for .t t

t

1 2( )kA tq

t

L

1 2tqL

tkA

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ch09 math textbook