Transcript Chapter 2

2.1 The Addition Property of
Equality
• A linear equation in one variable can be written in
the form: Ax + B = 0
• Linear equations are solved by getting “x” by
itself on one side of the equation
• Addition Property of Equality: if A=B then
A+C=B+C
• General rule: Whatever you do to one side of the
equation, you have to do the same thing to the
other side.
2.1 The Addition Property of
Equality
• Example of Addition Property:
x – 5 = 12
x = 17
(add 5 to both sides)
• Example using subtraction:
3
5
k  17  85 k
3
5
k  17  53 k  85 k  53 k
17  55 k  k
2.2 The Multiplication Property of
Equality
• Multiplication Property of Equality: if A=B and C
is non-zero, then AC=BC (both equations have the
same solution set)
• Since division is the same as multiplying by the
reciprocal, you can also divide each side by a
number.
• General rule: Whatever you do to one side of the
equation, you have to do the same thing to the
other side.
2.2 The Multiplication Property of
Equality
• Example of Multiplication Property:
3
4
h6
4
3
( 43 h )  43  6
h  43  16 
• Example using division
5x = 60
x = 12
(divide both sides by 5)
24
3
8
2.3 More on Solving Linear
Equations: Terms - Review
• As with expressions, a mathematical
equation is split up into terms by the +/-/=
sign:
2
1
1
3
x 
2
x 
6
x  3
• Remember, if the +/- sign is in parenthesis,
it doesn’t count:
2
3
x  3

1
2
x 
1
6
x  3
2.3 More on Solving Linear Equations:
Multiplying Both Sides by a Number
• Multiply each term by the number (using
the distributive property).
4x  3  42 x  1
4 x  12  8 x  4
• Within each term, multiply only one factor.
4x  3 y  1  42 x  1
4 x  12 y  1  8 x  4
Notice that the y+1 does not become 4y+4
2.3 More on Solving Linear
Equations: Clearing Fractions
• Multiply both sides by the Least Common
Denominator (in this case the LCD = 4):
1
1
 x  5 
x  3
4
2
4
4
 x  5 
x
4
2
 x  5  2 x  12
5  x  12
x  17
 43
2.3 More on Solving Linear
Equations: Clearing Decimals
• Multiply both sides by the smallest power of 10
that gets rid of all the decimals
.1 x  5  .05 x  .3
100  .1 x  5  100  .05x  100  .3
10 x  5  5 x  30  10 x  50  5 x  30
5 x  50  30
5 x  80  x  16
2.3 More on Solving Linear
Equations: Why Clear Fractions?
• It makes the calculation simpler:
1
1
 x  5 
x  1
47
94
94
94
 x  5 
x  94 1
47
94
2 x  5  x  94  2 x  10  x  94
x  10  94
x  104
2.3 More on Solving Linear
Equations
• 1 – Multiply on both sides to get rid of fractions
• 2 – Use the distributive property to remove
parenthesis
• 3 – Combine like terms
• 4 – Put variables on one side, numbers on the
other by adding/subtracting on both sides
• 5 – Get “x” by itself on one side by multiplying
both sides
• 6 – Check your answers
2.3 More on Solving Linear
Equations
• Example:
2
3
x  12 x  16 x  3
4x  3x  x  18
7 x  x  18
6 x  18
x3
2.4 An Introduction to Applications
for Linear Equations
• 1 – Decide what you are asked to find
• 2 – Write down any other pertinent information
(use other variables, draw figures or diagrams )
• 3 – Translate the problem into an equation.
• 4 – Solve the equation.
• 5 – Answer the question posed.
• 6 – Check the solution.
2.4 An Introduction to Applications
for Linear Equations
•
1.
2.
3.
4.
5.
6.
Find the measure of an angle whose complement is 10
larger.
x is the degree measure of the angle.
90 – x is the degree measure of its complement
90 – x = 10 + x
Subtract 10: 80 – x = x
Add x: 80 = 2x
Divide by 2: x = 40
The measure of the angle is 40 
Check: 90 – 40 = 10 + 40
2.5 Formulas - examples
•
•
•
•
•
•
A = lw
I = prt
P=a+b+c
d = rt
V = LWH
C = 2r
•
•
•
•
•
•
Area of rectangle
Interest
Perimeter of triangle
Distance formula
Volume – rectangular solid
Circumference of circle
2.5 Formulas
• Example: d=rt; (d = 252, r = 45)
then 252 = 45t
divide both sides by 45:
t 5
27
45
2.6 Ratios and Proportions
• Ratio – quotient of two quantities with the
same units
Examples: a to b, a:b, or
a
b
Note: percents are ratios where the second
number is always 100:
35
35% 
100
2.6 Ratios and Proportions
• Proportion – statement that two ratios are
equal
a
c
Examples: b
d

Cross multiplication:
if
a
b

c
d
then
ad  bc
2.6 Ratios and Proportions
• Solve for x:
81
x

Cross multiplication:
81 7  9  x
567  9  x
so x = 63
9
7
2.7 More about Problem Solving
• Percents are ratios where the second number is
always 100:
Example:
73
73%  100
 .73
If 70% of the marbles in a bag containing 40
marbles are red, how many of the marbles are
red?:
#red marbles =
70
100
 40  28
2.7 More about Problem Solving
• How many gallons of a 12% indicator solution
must be mixed with a 20% indicator solution to
get 10 gallons of a 14% solution?
Let x= #gallons of 12% solution,
then 10-x= #gallons of 20% solution :
10%( x)  20%(10  x)  14%(10)
10%( x)  20%( x)  2.0  1.4
 10%( x)  .6
x  ..61  6
2.8 Solving Linear Inequalities
•
•
•
•
<

>

means “is less than”
means “is less than or equal to”
means “is greater than”
means “is greater than or equal to”
note: the symbol always points to the
smaller number
2.8 Solving Linear Inequalities
• A linear inequality in one variable can be
written in the form:
ax < b (a0)
• Addition property of inequality:
if a < b then a + c < b + c
2.8 Solving Linear Inequalities
• Multiplication property of inequality:
– If c > 0 then
a < b and ac < bc are equivalent
– If c < 0 then
a < b and ac > bc are equivalent
note: the sign of the inequality is reversed when
multiplying both sides by a negative number
2.8 Solving Linear Inequalities
• Example:
 23 x  12 x  16 x  3
 4x  3x  x  18
 x  x  18
 2 x  18
x  9
-9