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Transcript
Section 4.1
Inequalities & Applications
 Equations

 3x
+ 7 = 13
 |y| = 7
 3x + 2y = 6
Inequalities
3x
+ 7 < 13
|y| > 7
3x + 2y ≤ 6

4.1
Symbols:

< > ≤ ≥ ≠
1
Overview of Linear Inequalities



4.1 Study Inequalities with One Variable
Why study Inequalities?
How do we show a Solution to an Inequality


With an Algebraic Statement -5 < x
By Graphing Using the Number Line
x ≤11
|x -1| < 3
(



Using Interval Notation (-5, ∞) (- ∞, 11] (-∞,-2)U(2,∞)
Via Set-Builder Notation { x | x > -5} {x | -4 <x ≤13 }
Use Properties to Solve Single-Variable Inequalities
4.1
2
Solving Inequalities






Examples of Inequalities containing a variable
-2 < a
x>4
x+3≤6
6 – 7y ≥ 10y – 4
6<x+2<9
A solution is a value that makes an inequality true
A solution set is the set of all solutions
Substitute a value to see if it is a solution:
Is 5 a solution to x + 3 < 6
no, 8 < 6 is false
Is -1 a solution to -3 > -5 – 2x no, -3 > -3 false
Is 2 a solution to y + 2 ≥ 4
yes, 4 ≥ 4 is true
4.1
3
Showing an Inequality Solution
on the Number Line

Identify the Boundary Point on the number line



Left ( or Right ) : the number is not a solution
Left [ or Right ] :
the number is a solution
Shade the number line over all solutions
)
4.1
4
Interval Notation


Is a compact way to precisely specify a complete solution set over a
range of numbers, without reference to a variable name
Parentheses ( ) and Brackets [ ] have another use






They are used to enclose a pair of values
The smallest number is always listed on the left
They have the same meaning as used in graphing: point included or excluded
Examples: (2, 5) [0, 15) (-4, -2] [-22, 43]
(-∞, 7) and [3, ∞) show how infinity values are noted
Interval notation can be used in addition to Set-builder
4.1
5
Example:
Number Line Graphs and Notation

What is the inequality for all real numbers greater than -5?


Either of the algebraic statements
x > -5 or -5 < x
Set-builder Notation (using the inequality) can also be used
{ x | -5 < x }


We can also show this interval in Interval Notation
as (-5, ∞)
We can graph this by identifying a portion of the number line.
Any portion of a number line is called an interval
For example, all real numbers greater than -5 up to ∞
is graphed
(
4.1
6
Unbounded Intervals
( or )
( point not included
[ or ]
[ point is included
(
a
[
a
)
a
]
a
a
Some textbooks use a hollow dot ○ in place of ( or )
and a solid dot
● in place of [ or ]
4.1
7
Principles Used to Solve Inequalities

Addition (and Subtraction) Principles

Any real number (or term) can be added to (or subtracted from)
both sides of an inequality to produce another inequality with the same
solutions.
The direction of the inequality symbol is unchanged.

a < b is equivalent to a + c < b + c
(true for >, ≥, < and ≤ )
Example 4 – Solve and graph
a) x + 5 > 1
x > -4 (-4,∞) {x| x>-4}
b) 4x – 1 ≥ 5x – 2
1 ≥ x (-∞,1] {x| x≤1}
4.1
8
Principles Used to Solve Inequalities

Positive Multiplication (and Division) Principles

If both sides of an inequality are multiplied (or divided) by a positive
number, the new inequality will have the same solutions.
The direction of the inequality symbol is unchanged.

a < b is equivalent to a • c < b • c
(true for >, ≥, < and ≤ )
Example 5 – Solve and graph
a) 3y < ¾
y < ¼ (-∞,¼) {y| y<¼}
b) ½t ≥ 5
t ≥ 10 [10,∞) {t| t≥10}
4.1
9
Principles Used to Solve Inequalities

Negative Multiplication (and Division) Principles



If both sides of an inequality are multiplied or divided by a negative
number, the new inequality will be equivalent
only if the direction of the inequality symbol is reversed !
Consider 5 < 7
Let’s multiply by -1: (-1)5 < (-1)7 … is -5 < -7 ?
No!
-5 > -7
so remember to reverse the symbol
Example 5 – Solve and graph
a) -5y < 3.2
y > -0.64 (-0.64,∞) {y| y>-0.64}
b) - ¼x ≥ -80
x ≤ 320 (-∞,320] {x| x≤320}
4.1
10
Principles Used Together

As in equations,


First use the addition principle to move variable terms to one side
and numeric terms to the other
Then use the multiplication principle to effect the solution
Example 6 – Solve and graph
a) 16 + 7y ≥ 10y – 2
y ≤ 6 (-∞,6] {y| y ≤ 6 }
b) -3(x + 8) – 5x > 4x – 12 x < 1 (-∞,1) {x| x < 1 }
4.1
11
Class Practice



Use the properties of linear inequalities to solve the following and
graph the solution set. Also give the solution in interval notation.
–2y + 6 ≤ –2
–2y ≤ – 8
y≥4
[
-9(h – 3) + 2h < 8(4 – h)
[4,∞)
9 95 ( x  3)  43 ( x  3)  x  1
 9h  27  2h  32  8h
5( x  3)  12( x  3)  9 x  9
 75hx 15
27 12
 32
 8h9 x  9
x  36

5
4
x  3  x  3  x  1
9
3
4.1
 51
 927
x 9 8h
 8h7x 27
 7x  9  7x  9
h 5
60  16 x
60
16
 154  x
12
Problem Solving

The statements
“a does not exceed b”
and
“a is at most b”
both translate to the inequality
a≤b

The statements
“a is at least b”
and
“a is no less than b”
both translate to the inequality
a≥b
4.1
13
Applications




Truck Rentals. Campus Fun rents a truck for
$45 plus 20¢ per mile. Their rental budget is
$75. For what mileages will they NOT
exceed their budget?
“for 150
____ miles or less” or
150
“mileage less than or equal to ____”
Charges = f(m) = 45 + .20m
45 + .20m ≤ 75
4.1
14
What Next?

Section 4.2 –
Intersections, Unions &
Compound Inequalities
4.1
15