Transcript Ch 3

Chapter 3
Linear Equations
3.1 Linear Equation in One Variable
A Linear Equation in one variable is an equation that
can be written in the form
ax + b = 0
Where a = 0
A linear function can be written as f(x) = ax + b
Examples of Linear equation.
2x –1 = 0, -5x = 10 + x, and 3x + 8 = 5
Properties of Equality
Addition Property of Equality
If a, b, c are real numbers, then
a = b is equivalent to a+ c = b + c.
Multiplication Property of Equality
If a, b, c are real numbers with c = 0, then
a = b is equivalent to ac = bc.
Example 8 Solving a Linear equation graphically
using graphing calculator
[ -6, 6, 1] by [-4, 4, 1]
Standard form of a line ( pg 158 - 159 )
An equation for a line is in standard form when it
is written as ax + by = c, where a, b, c are constants
With a, b and c are constants with a and b not
both 0
 To find x-intercept of a line, let y = 0 in the equation and
solve for x
 To find y-intercept of a line, let x = 0 in the equation and
solve for x
Example 115 ( Pg -163 )
[ 1984, 1991, 1] by [0, 350, 50] in 1987
In 1987 CD and LP record sales were both 107 million
3.2- Introduction to problem solving
Step1: Read the problems carefully to be sure that you
understand it. ( You may need to read the problem more than
once. ) Assign a variable to what you are being asked to find.
If necessary, write other quantities in terms of this variable.
Step 2: Write an equation that relates the quantities described in
the
problem. You may need to sketch a diagram, make a table, or
refer to
known formulas.
Step 3: Solve the equation and determine the solution.
Step 4: Look back and check your answer. Does it seem
reasonable? Did you find the required information ?
3.2 Solving a problem
The sum of three consecutive even integers is 96. Find
the three numbers.
Solution
Step 1: n is the smallest of the three integers
n + 2 : next consecutive even integer
n + 4 : larges of the three even integers
Step 2: Write an equation that relates these unknown quantities. As
the sum of these three even integers is 108
The equation is
n + (n + 2) + (n + 4) = 96
Step 3 : Solve the equation in Step 2
n + (n + 2) + (n + 4) = 96
( n + n +n) + (2 +4) = 96
3n + 6 = 96
3n + 6 = 96
3n = 90
n = 30
smallest of the three numbers is 30, so the three numbers are 30, 32, 34
Step 4: Check your answer. The sum of these three even integers is
30 + 32 + 34= 96
The answer checks.
Mixing acid ( Ex 12 pg 172)
20 %
2 liters
+
Step 1
x: liters of 60% sulphuric acid
x + 2: Liters of 50% sulphuric acid
Step 2
Concentration
0.20 (20%)
0.60 (60%)
0.50(50%)
60%
x liters
Solution Amount
2
x
x+2
=
Pure Acid
0.20(2)
0.60x
0.50(x + 2)
Equation
0.20(2)
+
0.60x =
0.50(x + 2)
(pure acid in 20% sol.) (pure acid in 60% sol.)
(pure acid in 50% sol.)
Step 3
Solve for x
0.20(2) + 0.60x = 0.50(x + 2)
2 (2) + 6x = 5(x + 2)
Multiply by 10
4 + 6x = 5x + 10 (Distributive Property) Subtract 5x and 4 from each side x = 6
Six liters of the 60% acid solution should be added to the 2 liters of 20% acid solution.
Step 4
If 6 liters of 60% acid solution are added to 2 liters of 20% solution, then there
will be 8 liters of acid solution containing 0.60(6) + 0.20(2) = 4 liters of pure acid.
Check
The mixture represents a 4/8 = 0.50 or 50% mixture
50%
x + 2 liters
Ex 50 (Pg 175) Anti freeze mixture
A radiator holds 4 gallons of fluid
x represents the amount of antifreeze that is drained and
replaced
The remaining amount is 4 – x
20 % of
+ 70 % of solution of = 50% of solution
4 – x gallons
x gallons
4 gallons
0.20(4-x) + 0.70x = 0.50(4)
0.8 – 0.2x + 0.7x = 2
0.5x = 1.2
x = 2.4
The amount of antifreeze to be drained and replaces is 2.4 gallons
Geometric Formulas
Perimeter of triangle P = a+b+c unit
Area of triangle
A = ½ bh sq.unit
a
c
h
b
Area of Rectangle =LW sq.unit
Perimeter P = 2(L + W) unit
W
L
Area of Parallelogram
A = bh sq.unit
h
b
Area & Volume of cylinder
A = 2 rh sq.unit
h
V = r2 h cu.unit
Area & Volume of a cube
r
A = 6a2 sq.unit
V = a3 cu.unit
a
a
a
3.3 Linear Inequality in One Variable
A linear inequality in one variable is an inequality
that can be written in the form
ax + b > 0, where a = 0. ( The symbol > may be
replaced with >, <, or < )
There are similarities among linear functions,
equations, and inequalities. A linear function is
given by f(x) = ax + b, a linear equation by ax + b
= 0, and a linear inequality by ax + b > 0.
3.3 Properties of Inequalities
Let a, b, c be real numbers.
 a < b and a+c < b+c are equivalent
( The same number may be added to or subtracted from both
sides of an inequality.)
 If c > 0, then a < b and ac < bc are equivalent.
(Both sides of an inequality may be multiplied or divided by the
same positive number)

If c< 0, then a < b and ac > bc are equivalent.
Each side of an inequality may be multiplied or divided by the
same negative number provided the inequality symbol is
reversed.
Examples of linear inequalities are
2x + 1< 0, 1-x > 6, and 5x + 1 < 3 – 2x
A solution to an inequality is a value of the variable that
makes the statement true. The set of all solutions is called
the solution set. Two inequalities are equivalent if they
have the same solution set.
Inequalities frequently have infinitely many solutions. For
example, the solution set to the inequality x- 5> 0 includes
all real numbers greater than 5, which can be written as
x > 5.
Using set builder notation, we can write the solution
set as { x x > 5 }.
Meaning
This expression is read as “ the set of all real numbers x such
that x is greater than 5. “
Ex -4 Graphical Solutions (Pg 182)
350
300
250
200
150
100
50
Distance
(miles)
y1
When x = 2 , y1 = y2, ie car 1 and car 2 both are 150 miles
From Chicago
y2
y2
y1
0
1
x=2
2
3
4
Time ( hours )
Distances of two cars
y1 < y2 when x< 2 car 1 is closer to Chicago than car 2
y1 is below the graph of y2
y1 > y2 when x > 2 Car 1 is farther from Chicago than Car 2
Y1 above the graph of y2
…Continued
Ex 5 (Pg 183)
Solving an inequality graphically
Solve 5 – 3x < x – 3
y1 = 5 – 3x and y2 = x – 3 Intersect at the point (2, -1)
5
4
3
2
1
-4
-3
-2
-1 0
-1
-2
-3
X=2
y1
y2
1
2
3
(2, -1)
4
y1 = y2 when x = 2
y1 < y2 when x > 2 , y1 is below the graph of y2
Combining the above result
y1 < y2 when x > 2
Thus 5 – 3x < x – 3 is satisfied when x > 2.
The solution set is {x / x> 2 }
Using Graphing Calculator
Y1 = 5 – 3x
Y2 = x - 3
Hit window
Hit Y
Enter inequality
Enter
[ -5, 5, 1] by [-5, 5, 1]
Hit Graph
Ex 92 ( Pg 178) Sales of CD and LP records
Using Graphing Calculator
Hit Y , enter equations
Enter table set
Enter window
Hit Table
Hit graph
1987 or after CD sales were greater than or equal to LP records
3.4 Compound inequalities
A compound inequality consists of two inequalities
joined by the words and or
In compound inequality contains word and, a solution must satisfy
both two inequalities.
Example 1
2x > - 5 and 2x < 3
2(1) > -5 and 2(1) < 3 1 is a solution
True
True
In compound inequality contains word or , a solution must satisfy
atleast one of the two inequalities.
Example 2
5 + 2 > 3 or 5 – 1 < -5
x + 2 > 3 or x – 1 < -5
True
False
5 is a solution
Symbolic Solutions and Number Lines
]
x<6
-8
-6
-4
-2
0
2
4
6
8
-6
(
-4
-2
0
2
4
6
8
x> - 4
-8
(
x < 6 and
x> -4
-8
-6
-4
-4<x<6
]
-2
0
2
4
6
8
Three-part inequality
(
-1 0 1 2 3 4
]
5
6
7
8 9
10
Numberline
4 < x < 10
Sometimes compound inequality containing the
word and can be combined into a three part inequality.
For example, rather than writing
x > 4 and x < 10
We could write the three-part inequality
4 < x < 10
Compound Inequality
Example
Solve
x + 2 < -1 or x + 2 > 1
x < -3
or x > -1 ( subtract 2 )
The solution set for the compound inequality results from taking the
union of the first two number lines. We can write the solution, using
Set builder notation, as
{ x x < - 3} U { x x > - 1 } or
{x x < - 3 or x > - 1}
x<-3
x>-1
x < - 3 or x > -1
-
-
-
)
-3
-4
-4
-4
-2
-3
)
-3
-1
(
-1
1
2
3

4

(
-1
-2
-2
0
0
1
2
3
4

0
1
2
3
4
Interval Notation
Inequality
Interval Notation
- 1<x<3
( - 1, 3)
-3<x<2
( - 3, 2]
-2<x<2
[ - 2, 2 ]
x < - 1 or x > 2
( -  , - 1) U (2,  )
x>-1
( - 1,  )
x<2
( - , 2 ]
Number line Graph
-4 -3
(
-4 -3
-
-2
-4 -3
-2
[
-2
-4 -3
-2
-4 -3
-2
(
-1 0 1 2
]
-1 0 1 2
]
-1 0 1 2
)
(
-1 0 1 2
(
-1 0 1 2
)
3 4
3 4
3 4
3 4
3 4
]
-
-4 -3
-2
-1 0 1 2 3 4
+
+
Example
Solve graphically and numerically. Write your answer in interval notation
x + 1 < -1 or x + 1 > 1
Y1 = -1, Y2 = x + 1, Y3 = 1
Enter y1, y2, y3
Enter window [ -5, 5, 1] by [ -5, 5, 1]

+1
0
Hit Graph
-2
-
-1

x + 1 < - 1 or x + 1 > 1
Solution in interval notation is ( - , - 2) U (0,  )
Example 6 (Page 195)
Solving a compound inequality numerically and graphically Using Technology
 Tution at private colleges and universities from 1980 to 1997
 Can be modelled by f(x) = 575(x – 1980) + 3600
 Estimate when average tution was between $8200 and $10,500.
Hit Y and enter equation
Hit Window and enter
Hit 2nd and Table
[ 1980, 1997, 1] by [3000, 12000, 3000 ]
Hit 2nd and calc and go to Intersect and enter 4 times to get intersection
School Enrollment
•
70
•
60
•
55
•
50
1970 1980
Year
1990
2000