Transcript CH2

Let c be any real number.
If a = b, then a + c = b + c
If a = b, then a ( c )= b ( c )
x + 7 – 7 = 22 – 7
x = 15
If a = b, then a – c = b – c
If a = b, then a /c = b /c
–7=–7
x = 15
Vertical steps = less writing.
2
+ 12 = + 12
x  41
5
5
x 9
2
x  26
“UNDO” the operations on the x term..
Direction: “SOLVE.” this means, WORK BACKWARDS
Rules to follow:
SIMPLIFY to 1 variable term. You may need to combine like
terms, distributive property to remove ( )’s, and Clear Fractions by
multiplying both sides by the LCD (least common denominator).
3x  5  17
–5=–5
3x  12
3
3
x4
45  x  13
– 45
= – 45
 x  32
-1
-1
x  32
4
x 7 1
3
+7 = +7
2
4
x 8
3
3
4
3
4
x6
16.3  7.2x  8.18
– 16.3
= – 16.3
 7.2x  24.48
-7.2
-7.2
x  3.4
Sect 2.2
Solving using Order of Operations
Combining Like Terms in the simplifying stage.
3x  4x  14
7 x  14
7
7
x  2
Always move the smallest variable term!
 x  5  8x  6
+ 8x = + 8x
7x  5  6
–5=–5
7x  1
7
7
1
x
7
6x  5  7 x  10  4x  7
5 1x  17  4x
+ 4x = + 4x
5  3x  17
–5
=–5
3x  12
3
3
x4
Distributive Property in the simplifying stage.
2  5x  5  3x  2 1
2  5x  25  3x  6 1
 5x  23  3x  7
+ 5x
=+ 5x
 23  8x  7
+7= +7
16  8x
8
8
2  x
To clear fractions, multiply both sides by the LCD.
LCD = 6
2
2
16
6 x   2x 6
3
6
4x 1  12x
– 4x
= – 4x
1 8x
8
8
1
x
8
Two approaches.
2
3x  2  8
5
6
4 5
5
x 8 5
5
5
6x  4  40
–4=–4
6x  36
6
6
x6
5
2
4
2
3x  2  8 5
5
2
3x  2  20
–2=–2
3x  18
3
3
x6
To clear fractions, multiply both sides by the LCD.
6
12
6
12
3
3 12
LCD = 12
1
3
2 4
x    12
2
2 4 3
6x  18  9  8
1005.3  1.2 x  1.94100
530  120x  194
6x  9  8
– 530
= – 530
120x  336
–9=–9
6x  1
To clear decimals, multiply both sides by 10.
Every multiplication of 10 moves the decimal
one place to the right. We will multiply by 100.
x
1
6
120 = 120
x  2.8
6
6
All the above examples are Conditional equations…they have a solution.
Contradiction Equation. (No solution) Identity Equations. (Have infinite sols.)
2x  7  7x  1  5x
3x  5  2x  2  x
3x  5  2x  4  x
2 x  7  7 x  7  5x
3x  5  3x  4
2x  7  2x  7
 5  4
77
False Statement
No Solution
Contradiction Equation
True Statement
Infinite Solutions
Identity Equation
1
A  bh
2
1
2  A  bh  2
2
Solve for b.
Isolate the b by
undoing the
operations
taking place on
the variable b.
2 A  bh
h
h
2A
b
h
Solve for W.
Remove the terms
with no W 1st.
Next isolate the W
by undoing the
operations taking
place on the
variable W.
P  2L  2W
– 2L = – 2L
P  2L  2W
2
2
P  2L
W
2
Solve for a. K  917  6w  h  a
– 917 = – 917
Move the terms
and factors that
are outside the
( ) ‘s to the
other side.
Since we have
– a, add the a
to the left and
subtract the
fraction to the
right. a will
be isolated.
K  917  6w  h  a
6
6
K  917
 wha
6
a
K  917
K  917

a
6
6
a  wh
K  917
6
Solve for h.
Remove the terms
with no h 1st. Next
isolate the h by
undoing the
operations taking
place on the
variable h.
A  2rh  2r 2
 2r 2 
 2r 2
A  2r 2  2rh
2r
2r
A  2r 2
h
2r
Sect 2.4
Solving Percentage Problems
n
 1 

n
n% =

  n0.01
100
 100 
Convert decimals (fractions) to percent and percent to decimal.
Fraction to a percent
n
%

d 100
Convert to a percent.
3
x

5
100
Cross multiply and
products are equal.
5x  300
5 x 300

5
5
x  60
60%
Decimal to a percent.
Multiply by 100
Convert to a percent.
0.359 100  35.9%
Notice we moved the
decimal point 2
places to the right.
Percent to a decimal.
Drop % and divide
by 100
Convert to a decimal.
19.9%
19.9 100  0.199
Notice we moved the
decimal point 2
places to the left.
The values
that associates
with the
words “is”
and “of”
We will work both techniques…pick your favorite.
x  11 0.01 49
x  5.39
Not to bad… just like Sect 1.1 and 1.2.
What is 11% of 49?
x
11  is
100 49
of
100x  11 49
100 100
x  5.39
Notice we will
always cross
multiply by the
numbers on the
diagonal and
divide by the
3rd number!
3  16.01  x
3  0.16 * x
0.16 0.16
18.75  x
is
16  3
100
x
of
Cross multiply the numbers on the diagonal
and divide by the 3rd number! Calculator !!!
x  100  3 16  18.75
x  .01  50  34
Here is where problems may occur!
x really needs to multiplied by 0.01
because of the word percent!
0.50x  34
0.50 0.50
x  68
68%
x  34
is

100 of
50
Cross multiply the numbers on the diagonal
and divide by the 3rd number! Calculator !!!
x  100  34  50  68
68%
In 2006, there were 300 million people in the United States, and 62.2% of them
lived within 5 mi. of a Wal-Mart store. How many people lived with 5mi. of a
Wal-Mart store?
Any number that represents a TOTAL associates with 100% and is
across from 100 in the proportion or associates with “of.”
62.2 is
x

100 300
of
x  62.2  300 100  186.6 million
About 1.6 million students who graduated high school went to college. This was
66% of all high school graduates. How many total high school graduates are
there?
TOTAL is across from 100 in the
proportion or associates with “of.”
66
100

1is.6
of
x
x  100 1.6  66  2.42 million
A car dealer lowered the sticker price of a car from $20,830 to $18,955.
What percent of the regular price does the sale price represent?
What is the percent discount? Any number that represents a ORIGINAL PRICE associates with 100%
and is across from 100 in the proportion or associates with “of.”
x
18955
is
x  100 18955  20830  90.563784
100 20830
of
 90.6%
What is the percent discount? 100%  90.6%  9.4%

The total bill was $47.70 that included 6% sales tax. How much was the
merchandise before tax?
We have TOTAL and ORIGINAL price…which is the “of”?
106
6
100

47
is.70
ofx
ORIGINAL price is the “of”? The $47.70 represent the ORIGINAL
price + 6% sales tax
WAIT a minute! That means our PERCENTAGE is not 6%, 106%!
x  100  47.70 106  45
$45
Sect 2.5 Word Problems
Sect 2.5 Word Problems
Sect 2.5 Word Problems
10x  2  78
 2  2
10 x  80
10

10
53  2 x  70
15  10x  70
15
 15
x  5.5
10 x 55
10
10
24  3x  34
8  6x  34
8
x 8
 8
6x 
 26 x  26  13
6
3
6
6
Sect 2.5 Word Problems
260km
3 times the unknown
distance = 3x
P
How far he biked = 3x
3  65  195km
How far he had left to go = x
x  65km
3x
Unknown distance = x
S
3x  x  260
4x  260
x  65
x
F
Sect 2.5 Word Problems
Mile
x
Mile
x+1
1 mile
First marker = x
x  279
Next consecutive
marker = x + 1
279  1  280
x + x + 1 = 559
2x  1  559
2x  558
x  279
Sect 2.5 Word Problems
Length
P  2L  2W
288  2L  2W
288  2L  2W
Width
 W  50 ft
Width
288  2W  44  2W
288  2W  88  2W
 L  W  44
Length
 L  W  44
 50  44  94 ft
288  4W  88
200  4W
50  W
W
94 ft by 50 ft
Sect 2.5 Word Problems
B = Number of brochures to print
Starting Cost
2($300)
+ The number of Brochures
+ B(the cost of each Brochures)
Change cents to dollars!
$600
 $3000
+ B($0.215)
 $3000
 $3000
0.215B  2400
B  11162.8
Can print 11,162 brochures and not go over budget.
Sect 2.5 Word Problems
Back side is the unknown = x
 40
 2  40  80
Front = x + 20  40  20  60
Peak = 2x
The sum of three angles of a
triangle is 180 degrees.
x + 2x + x + 20 = 180
4x + 20 = 180
4x = 160
x = 40
40  80  60  180
Sect 2.5 Word Problems
H = the hammer price of final bid.
$1150 + 8%(of final bid) = final bid
1150  0.08H  1H
1150  0.92H
1250  H
The final bid must be $1250 or higher.
Proportion Style.
Jared has to pay the Auctioneer
8% of the final bid, so he gets
92% of the final bid.
is
92  1150
100
ofH
H  100 1150  92  1250
Let A > B and C is a non-zero positive constant.
AB
 C  C
AC  B C
AB
 C  C
AC  B C
REMEMBER!!!
AB
 C  C
AC  B C
AB
 C  C
AC  B C
A B

C C
  
  
    
   
  
How to write a solution set with set builder notation, interval notation
and Graph.
Let x > 3 be our solution.
Graph.
Set builder notation
“The set of all x, such that x > 3.”
(
0
{x|x>3}
Interval notation
3 Author.
3 Mr. Fitz
3,
Let x < 3 be our solution.
Author.
x | x  3
]
Mr. Fitz
Interval notation


,3
0 3
REMEMBER!
3
When x is on the left side the inequality symbol points in the direction of the graph.
Compound inequalities
Let x > – 2 and x < 3 be our solution. We want all x values between -2 and 3.
(
]
2 x x 3
x | 2  x  3
2 0 3
Author.
Interval notation
2
 2,3
3
Mr. Fitz
Solve the inequalities. Graph the solution and write in interval notation.
3x 1  2x  5
– 2x
– 2x
x 1  5
+1
+1
x  4
4
 , 4
4
1
x7
4
4
 2 x  18
–2
–2
x  9
x  28
28
 , 28
Flip
inequality
symbol
9
 9,  
6  5x  7
–6
–6
 5x  1
–5
–5
1
x
5
2x  9  7 x  1
– 2x
– 2x
If you move the smallest variable
term, it will stay positive.
 9  5x  1
–1
1
5
 ,  15 
3x  27 1  2  5x  30
3x  28  5x  28
–1
10  5x
5
x  2
2
+ 5x
+ 5x
8x  28  28
+ 28
5
2 x

3x  9 1  2  5x  6
Switch the
inequality
around so x is
on the left side.
 2,  
+ 28
8x  0
8
8
x0
0
 , 0
Let p = how many people
$50
+ $15(per person)
50  15 p  450
– 50
15 p  400
– 50
15
15
< $450
p  26.6
The party cannot exceed 26 people!
Let h = how many hours worked in the 4th week
Average of 4 wks > 16 hours/week
4
20  12  14  h
 16
4
4
46  h  64
– 46
– 46
h  18
Dina has to work at least
18 hours in the 4th week.