Math 095 Chapter 7 Chapter 7 - Amazon S3 · 2016-06-22 · 7.1 Simplifying Rational Expressions Math 095 Chapter 7 → Simplify rational expressions → Factors that are opposite

Math 095 Chapter 7

Chapter 7

7.1 Simplifying Rational Expressions7.2 Multiplying and Dividing Rational Expressions7.3 Addition and Subtraction with Like Denominators;Least Common Denominators7.4 Addition and Subtraction with Unlike Denominators7.5 Simplifying Complex Fractions7.6 Solving Rational Equations7.7 Problem Solving Using Rational Equations7.8 Proportions and Similar Triangles (SKIP)

1 of 31

7.1 Simplifying Rational Expressions Math 095 Chapter 7

→ Simplify rational expressions→ Factors that are opposite in sign (i.e. that reduce to a −1)

Is this always definedx + 2

x + 1?

The domain of an algebraic expression is the set of all real numbers for which theexpression is defined.

Reducingab

ac=

b

c

Example:6x2

12x3=

6x2

2 • 6 • x2 • x=

1︷︸︸︷

✟✟6x2

2 •1

︷︸︸︷

✟✟6x2 •x

=1

2x

36x4

9x7=

Can I reduce the following this way?

x + 2

x + 3=

✚x + 2

✚x + 3=

2

3

x2 − 5

−5=

x2 − ✁5

−✁5= x2

x2 + 5x + 3

x2 + 5x=

✘✘✘✘✘x2 + 5x + 3

✘✘✘✘✘x2 + 5x

= 3

2 of 31


Set Builder Notation

{x|x ∈ R, }

{ x | x ∈ R }The set x such that x is a real number (conditions go here)

http://www.purplemath.com/modules/setnotn.htm

Other Types of Numbers Notation.

N the counting numbersZ the integersQ the rational numbers (fractions)R the real numbersI the irrational numbersC the complex numbers

You can only reduce FACTORS!

Example:

✘✘✘✘(x + 3)(x − 3)

✘✘✘✘(x + 3)(x − 1)=

(x − 3)

(x − 1)

Restriction on x?

Domain:

x2 + 5x + 6

x2 − 9=

Restriction on x?

Domain:

x2 − 4

x2 + 4x + 4=

Restriction on x?

Domain:

3 of 31


Considerx − y

y − x=

(x − y)

(y − x)• −1

−1=

−1 • (x − y)

(−y + x)=

−1 •✘✘✘✘(x − y)

✘✘✘✘(x − y)= −1 • 1 = −1

4 of 31

7.2 Multiplying and Dividing Rational Expressions Math 095 Chapter 7

Recall:5

3• 6

25=

1︷︸︸︷

✁51

︷︸︸︷

✁3

•

2︷︸︸︷

✁65

︷︸︸︷

✚✚25

=2

5

Example 1)

5x4

6x• 2

x=

5x4 •1

︷︸︸︷

✁23

︷︸︸︷

✁6 x • x

=5x4 • 1

3x2=

5x4−2

3=

5x2

3{x|x ∈ R, x 6= 0}

Example 2)

x2 − 25

x2 − 4x + 3÷ x + 5

x − 3=

x2 − 25

x2 − 4x + 3• x − 3

x + 5

=(x − 5)✘✘✘✘(x + 5)

(x − 1)✘✘✘✘(x − 3)• ✘✘✘x − 3

✘✘✘x + 5

=x − 5

x − 1{x|x ∈ R, x 6= 1, 3, −5}

You try

1)10

x7• 3x2

25x=

2)x2 − 1

x÷ x + 1

x − 1=

3)x − 6

6÷ 6 − x

3=

4) (y2 − 9) ÷ y2 − 2y − 3

y2 + 1=

5 of 31

7.3 Addition and Subtraction with Like Denominators; LCD Math 095 Chapter 7

Recall to add fractions we need a common denominator

1

5+

2

5=

1 + 2

5=

3

5

3

x+

9

x=

3 + 9

x=

12

x

Domain:

y2 + 5y

y + 3− 4y + 6

y + 3=

y2 + 5y − (4y + 6)

y + 3=

y2 + 5y − 4y − 6

y + 3=

y2 + y − 6

y + 3

=✘✘✘✘(y + 3)(y − 2)

✘✘✘y + 3{y|y ∈ R, y 6= −3}

x

15+

2x + 5

15=

y2 + 6y

y + 2+

2y + 12

y + 2=

y2 − 7y

y2 + 8y + 16+

6y − 20

y2 + 8y + 16=

x2 + 9x

x + 7− 4x + 14

x + 7=

5x2 + 4x

x − 1− 6x + 3

x − 1=

6 of 31

7.4 Addition and Subtraction with Unlike Denominators Math 095 Chapter 7

LCD = Least Common Denominator (Similar to Least Common Multiple)

Recall1

2+

1

3+

1

5=

LCD = 2 • 3 • 5 = 30

1

2+

1

3+

1

5=

30+

30+

30=

Now let’s add some variables

1

x+

1

x + 1=

LCD =

1

x + 2+

1

x + 3=

LCD =

1

2x− x

x + 1=

LCD =

Now let’s factor so we can reduce

6x + 18

x2 + 3x=

7 of 31

7.4 Addition and Subtraction with Unlike Denominators Math 095 Chapter 7

4x2 + 4x + 1

1 − 4x2=

12x4 − 18x3 − 12x2

12x3 + 42x2 + 18x=

1

(x + 1)(x + 2)+

1

(x + 2)(x + 3)=

LCD =

1

x2 + 2x + 1− 1

x2 + 3x + 2=

LCD =

8 of 31

7.5 Simplifying Complex Fractions Math 095 Chapter 7

Complex Fractions - fractions within fractions

Example 1

1214

=1

2÷ 1

4=

1

2• 4

1= 2

Write all fractions in improper fractional form, if they are a mixed number, BEFOREstarting the simplification process!

LEAVE ALL ANSWERS AS EITHER A REDUCED FRACTION (PROPEROR IMPROPER) OR AS AN INTEGER. NO MIXED NUMBERS! NO DECI-MALS!

Example 2

323

116

=11376

=11

3÷ 7

6=

11

3• 6

7=

11 •2

︷︸︸︷

✁61

︷︸︸︷

✁3 •7

=11 • 2

1 • 7=

22

7

Example 3

x852y

=x

8÷ 5

2y=

x4

︷︸︸︷

✁8

•

1︷︸︸︷

✁2 y

5=

xy

20

Example 4

2p2p+5

14p+10

=2p

2p + 5÷ 1

4p + 10=

2p

2p + 5• 4p + 10

1=

2p

✘✘✘✘✘(2p + 5)• 2✘✘✘✘✘(2p + 5)

1= 4p

2 methods of solving Complex Fractions1) Simplify the numerator and denominator with separate LCD.2) Use 1 LCD to clear ALL fractions.

9 of 31

7.5 Simplifying Complex Fractions Math 095 Chapter 7

Method 1 Method 21

x−3 − 1x+3

1 − 1x2−9

1x−3 − 1

x+3

1 − 1x2−9

(x+3)(x−3)(x+3) − (x−3)

(x−3)(x+3)

x2−9x2−9 − 1

x2−9

1x−3 − 1

x+3

1 − 1x2−9

x2−91

x2−91

(x+3)−(x−3)(x−3)(x+3)

x2−9−1x2−9

x2−9x−3 − x2−9

x+3

(x2 − 9) − x2−9x2−9

6(x−3)(x+3)

x2−10x2−9

(x + 3) − (x − 3)

x2 − 9 − 1

6

(x − 3)(x + 3)• x2 − 9

x2 − 10

6

x2 − 10

6

x2 − 10

14x2 − 1

x3

1x−1 + 1

x−1

=

10 of 31

7.6 Solving Rational Equations Math 095 Chapter 7

→ Solving Rational Equations (i.e. find a solution)

Recall

Expressions Equations→ No equal sign → Equal sign→ Simplify → Solve→ Rewrite

What can we do to Rational Expressions?1) Factor and Reduce (Rewrite-Simplify)2) Multiply and Divide rational expressions (factor and reduce) (LCD)3) Add/Subtract rational expressions (factor and reduce) (Rewrite-Simplify)4) Simplify complex rational expressions

→ Use all previous skills→ Factor and reduce

Solve a new type of equation - a rational equation

Procedure to solve a rational equation.1) List any restrictions that exist (i.e. what is the domain - no division by zero, etc.)2) Multiply each side by the LCD3) Solve4) Check (if necessary).

Example 1:2

3+

5

6=

9

x

LCD =

Domain =

11 of 31


Example 2:x

x − 2+

2

3=

2

x − 2

LCD =

Domain =

1

2x=

2

x− 3

8

LCD =

Domain =

3x

2+

4

x= 5

LCD =

Domain =

12 of 31


x

x − 2+

1

x − 4=

2

x2 − 6x + 8

LCD =

Domain =

1

x − 1− 1

x + 1=

3x

x2 − 1

LCD =

Domain =

2

1 + x=

3

1 − x+

5

x

LCD =

Domain =

13 of 31


3

2x − 6− x + 1

4x − 12= 4

LCD =

Domain =

14 of 31


Solving formula - Solving literal equations.

F =9

5C + 32 Solve for C.

K =2(ab − c)

fgSolve for c.

1

R=

1

r1

+1

r2

Solve for r1.

P = 2x + 2y Solve for y.

S = 2LW + 2LH + 2WH Solve for W .

D = rt Solve for t.

15 of 31


t

4+

t

5= 1 Solve for t.

T = mg + mf Solve for m.

1

p+

1

q=

1

fSolve for f .

ab = ac + d Solve for a.

16 of 31

7.7 Problem Solving Using Rational Equations Math 095 Chapter 7

Problem Types→ Number→ Work→ Distance

17 of 31


Logistics of Solving a Story Problem (Application Problem)

How to solve a story problem1) Read it and understand what is being asked.

a) Jot down important facts - in your own words.

2) Define a Variable that represents the solution to the question asked.3) Use a table or data format that allows you to gather and organize the data.4) Use the table (data format) to write an equation.5) Solve the equation.6) Answer the question!! (Complete sentence.)

Example ) A Number, minus five times its reciprocal, is 4. Find the number.

Let x = the number

then1

x= the reciprocal

x − 5(

1

x

)

= 4

x2 − 5 = 4x

x2 − 4x − 5 = 0(x + 1)(x − 5) = 0x = −1, 5

18 of 31


Example 1)

Suppose one painter can paint an entire fence in 5 hours, and a second painter takes 3hours. How many hours would it take to paint the fence together?

Let t = time together

Work Area Time Alone Rate Alone Time Together Work Together

Painter 1 51

5t

1

5t

Painter 2 31

3t

1

3t

Number of jobs 1

1

5t +

1

3t = 1

3t + 5t = 158t = 15t = 1.875 hours = 1 hours 52.5 minutes

Question id: 27924

19 of 31


By checking work records, a carpenter finds that Juanita can build a small shed in 12hours. Anton can do the same job in 16 hours. How long would it take to build 5 sheds ifthey worked together?

Let (Declare your variable here.)


Juanita

Anton

Number of jobs

20 of 31


One painter can paint a house in 10 hours, yet a more experienced painter can paint thehouse in 8 hours. How many hours would it take the two painters, working together, to paintthe house? (If needed, round minutes to 1 decimal place.)



Painter 1

Experienced Painter

Number of jobs

Tom can do a job in 120 minutes; Tom and Jerry can do the job together in 48 minutes.How long for Jerry to do the job by himself?



Tom

Jerry

Number of jobs

21 of 31


Ben takes 3 hours to wash 750 dishes, and Frank takes 4 hours to wash 450 dishes. Howlong will they take, working together, to wash 2000 dishes?



Ben

Frank

Number of jobs

It takes Jim 2 hours to put up paneling on 1 wall in a room. Omar takes 3 hours to dothe same job. How long would it take them to panel 4 walls?



Jim

Omar

Number of jobs

22 of 31


a) The flow rate for one pipe is 11

4times that of another pipe. A swimming pool can be

filled in 4 hours using both pipes. Find the time required to fill the pool using only the pipewith the lower flow rate.



Fast Pipe

Slow Pipe

Number of jobs

b) Assume that the pipe with the higher flow rate above is shut off after 1 hour and ittakes an additional 10 hours to fill the pool. Find the rate of each pipe.


Work Area Time Alone Rate Alone Time Together Work TogetherFast Pipe

Slow Pipe

Number of jobs

23 of 31


Solving quadratics with the Quadratic Formula - a preview

Given ax2 + bx + c = 0 then x =−b ±

√b2 − 4ac

2a

Why should we use the quadratic Formula?

Example 1) Solve 0 = 3x2 − 16x − 12 using the quadratic formula.

a = 3b = −16c = −12

Substitute into the quadratic formula

x =−b ±

√b2 − 4ac

2a=

−(−16) ±√

(−16)2 − 4(3)(−12)

2(3)=

16 ±√

256 + 144

6

x =16 ±

√400

6=

16 ± 20

6

x =16 + 20

6or x =

16 − 20

6

x =36

6or x =

−4

6

x = 6 or x = −2

3

You try: 0 = 4x2 − 20x − 144

24 of 31


You try: 0 = 3x2 − 24x − 27

Uniform Motion

Distance=Rate•Time d = r • t

Same direction |——>——>|Towards each other |——><——|

Equal Time |—->||——>|

Equal Distance |——>||<——|

25 of 31


Example 1)

A boat, which moves at 20 miles per hour in water without a current, travels 3 milesdownstream in the same amount of time it takes to travel 2 miles upstream. Find the speedof the current.

Let x = the speed of the current.

Work Area Distance Rate Time

Upstream 2 (20 − x)2

(20 − x)

Downstream 3 (20 + x)3

(20 + x)

2

(20 − x)=

3

(20 + x)

2(20 + x) = 3(20 − x)40 + 2x = 60 − 3x

5x = 20x = 4

Question id: 27947

The speed of the current is 4 miles per hour.

26 of 31


A boat, which moves at 18 miles per hour in water without a current, travels 14 milesdownstream in the same amount of time it takes to travel 10 miles upstream. Find the speedof the current.



Upstream

Downstream

The speed of the current is miles per hour.

27 of 31


Example 2)

A boat, which moves at 4 miles per hour in water without a current, travels 12 milesdownstream and travels 12 miles upstream in 8 hours. Find the speed of the current.

Let x = the speed of the current.


Upstream 12 (4 − x)12

(4 − x)

Downstream 12 (4 + x)12

(4 + x)

Total 8

12

(4 − x)+

12

(4 + x)= 8

12(4 + x) + 12(4 − x) = 8(4 + x)(4 − x)48 + 12x + 48 − 12x = 8(16 − x2)96 = 8(16 − x2)12 = 16 − x2

x2 = 4x = ±2

Question id: 27957

The speed of the current is 2 miles per hour.

28 of 31


A boat, which moves at 10 miles per hour in water without a current, travels 15 milesdownstream and travels 15 miles upstream in 4 hours. Find the speed of the current.



Upstream

Downstream

Total

The speed of the current is miles per hour.

29 of 31


Example 3)

The current of the river is 2 miles per hour. If a boat goes 8 miles upstream and 8 milesback again in 3 hours. Find the speed of the boat in water without a current.

Let x = the speed of the boat.


Upstream 8 (x − 2)8

(x − 2)

Downstream 8 (x + 2)8

(x + 2)

Total 3

8

(x − 2)+

8

(x + 2)= 3

8(x + 2) + 8(x − 2) = 3(x + 2)(x − 2)8x + 16 + 8x − 16 = 3(x2 − 4)16x = 3x2 − 120 = 3x2 − 16x − 12x = 6.0000, −0.6667

Question id: 27960

The speed of the boat is 6 miles per hour.

30 of 31


The current of the river is 6 miles per hour. If a boat goes 10 miles upstream and 10miles back again in 4 hours. Find the speed of the boat in water without a current.

Let


Upstream

Downstream

Total

The speed of the boat is miles per hour.

31 of 31

Documents

Math 095 Chapter 7 Chapter 7 - Amazon S3 · 2016-06-22 · 7.1 Simplifying Rational Expressions Math 095 Chapter 7 → Simplify rational expressions → Factors that are opposite