Before we begin, let's remember the primary rule for working with linear inequalities:
... when you multiply (or divide) an inequality by a negative value,
you must change the direction of the inequality.

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Now, let's see what happens when inequalities work in conjunction
(together) with one another.

A compound inequality is two simple inequalities joined by "and" or "or".

Compound Inequalities : OR
holdOR
Find the values for x which satisfy:
2x + 3 < 7 or 5x + 5 > 25

Solve each inequality separately.
Combine the results using the word "or".

Solve first inequality:
2x + 3 < 7
2x < 4
x < 2
Solve second inequality:
5x + 5 > 25
5x > 20
x > 4
The solution is x < 2 or x > 4.
Interval notation: OR
ORgraph2

The combined shaded regions on the number line represent numbers that are solutions of either x < 2 or x > 4 (or both if possible). These combined regions are referred to as the union of the two inequalities.

Example 1:
Solve for a:     a - 5 > 0   or   3a - 1 < 8

a - 5 > 0
a > 5

3a - 1 < 8
3a < 9
a < 3
The solution is a < 3 or a > 5.
ORgraph3
The answer is all real numbers that are less than or equal to 3 OR greater than 5.


Example 2: 
Write a compound inequality shown by this graph:
                                         Orgraph4

Let variable = x
The left side:
x < -1
The right side:
x > 2


The solution is:
x < -1 or x > 2

This problem is OR as the graphs to not overlap.


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Compound Inequalities : AND
AND girl
Find the values for x which satisfy:
8 < 3x - 1 and 4x + 2 < 22

Solve each inequality separately.
Combine the results using "and".
Look for where the solutions overlap.

Solve first inequality:
8 < 3x - 1
9 < 3x
3 < x
Solve second inequality:
4x + 2 < 22
4x < 20
x < 5
The solution is x > 3 and x < 5.
Can also be written as: 3 < x < 5
Interval notation: (3, 5)
andgraph1

When working with AND, you are looking for an overlapping region. You are looking for the numbers that are solutions for BOTH x < 3 and x > 5 at the same time. This overlapping region is referred to as the intersection of the two inequalities.

Example 3:
Solve for a:     3a - 9 < 12   and   3a - 9 > -3

3a - 9 < 12
3a < 21
a < 7

3a - 9 > -3
3a > 6
a > 2
The solution is a < 7 and a > 2.
andgraph2
Answer is the overlapping of the two inequalities.
andgraph3


Example 4:
Solve for x:     -10 < 2x + 4 < 30
This solution can be approached in two ways.
1) You can deal with it as it appears and solve through the inequalities at the same time.
2) You can separate the problem into two parts and solve each part separately.
Either way, you will arrive at the same answer.

Method 1:
-10 < 2x + 4 < 30
subtract 4 from all sections
-14 < 2x < 26
divide all section by 2
-7 < x < 13
Method 2:
-10 < 2x + 4       and         2x + 4 < 30
-14 < 2x                            2x < 26
-7 < x                                 x < 13
            -7 < x   and   x < 13

 

Example 5: 
The antifreeze added to your car's cooling system claims that it will protect your car to -35º C and 120º C.  The coolant will remain in a liquid state as long as the temperature in Celsius satisfies the inequality  -35º < C < 120º.  Write this inequality in degrees Fahrenheit.

caryelling
Remember:
lineq10
Plan:
-- set up inequality
--
substitute for C
-- solve for (isolate) F

linqu11
Solution: The coolant will remain in a liquid state as long as the temperature in Fahrenheit degrees satisfies the inequality
-31º < F < 248º


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Applied Problem Using AND and OR: 

The height of a horse is measured in a vertical line from the ground to the withers (at the base of the neck).  Horses are measured in "hands" where one hand = 4 inches.  If a horse is more than an exact number of hands high (hh), the extra inches are given after a decimal point, e.g. 14 hands 2 inches is written as 14.2 hh.  Normal riding horses are between 14.3 hh and 17 hh, inclusive.  Horses shorter than 14.3 hands are called ponies and horses over 17 hh are often called draft (or work) horses.  

a)
  Write an inequality statement to represent the heights of normal riding horses in inches.

b)  Write an inequality statement stating the heights, in inches, of equines (horses) that do not fit the normal riding horse height specifications.
horsewithers

Solution:
a)  Normal riding horse heights in hands: 
14.3 hh < h < 17 hh
Convert to inches.
14.3 hh = 14(4) + 3 inches
             = 59 inches
17 hh = 17(4) inches
          = 68 inches

Answer:  Normal riding horse height in inches:
59" < h < 68"

b)  Equines outside of the normal riding horse heights in hands:
h < 14.3 hh  or  h > 17 hh
Use conversions
from part a.
Answer:  Equine heights in inches not fitting the normal riding horse heights:
h < 59"  or   h > 68"

 

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hint gal
Yes, there is a way to determine solutions for inequalities on your graphing calculator. Click the calculator at the right to see how to use the calculator with single variable inequalities.
ti84c
For calculator help with inequalities
(single variable)

click here.


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