Linear Inequalities

Remember the primary rule for working with linear inequalities:
... multiplying or dividing by a negative number
changes the direction of the inequality.

 

Inequalities with one variable:

In Math A you worked with Linear Inequalities.  You will now be extending that knowledge to more sophisticated problems and more applied situations.
Click here to review "the basics" of one-variable Linear Inequalities from Math A.
(opens in new window)
 

You will now be seeing more references to interval notation when working with linear inequalities.  Check the table below if you need a quick review of this notation.

How to use your TI-83+ graphing calculator with one variable inequalities.
Click calculator.
 Interval Notation:  (description)

(diagram)
Open Interval:   (a, b)  is interpreted as a < x < b  where the endpoints are NOT included.
(While this notation resembles an ordered pair, in this context it refers to the interval upon which you are working.)

(1, 5)
Closed Interval:  [a, b]  is interpreted as a < x < b  where the endpoints are included.

[1, 5]
Half-Open Interval:  (a, b]  is interpreted as a < x < b  where a is not included, but b is included.

(1, 5]
Half-Open Interval:  [a, b) is interpreted as a < x < b where a is included, but b is not included.

[1, 5)
Non-ending Interval:  is interpreted as x > a where a is not included and infinity is always expressed as being "open" (not included).


Non-ending Interval:  is interpreted as x < b where b is included and again, infinity is always expressed as being "open" (not included).

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

Solving an "And" Compound Inequality:

3x - 9 < 12 and 3x - 9 > -3

Also written ... 
         

Or written ...

Isolate the variable between the two inequality signs
(or solve each side separately.)

The solution is 2 < x < 7,
which can be read x > 2 and x < 7.
Interval notation:  [2, 7]


Solving an "Or" Compound Inequality:

2x + 3 < 7  or  5x + 5 > 25

Also written ...
      [2x + 3 < 7]    [5x + 5 > 25]

Solve the first inequality
  Solve the second inequality
The solution is x < 2 or  x > 4.
Interval notation:  


 
Applied Problem Using "AND"
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.
 
Solution:
Remember:

Plan:
-- substitute for C
-- solve for (isolate) F

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


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.
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"

 

Inequalities with two variables:


Linear Inequalities may also appear with two variables, such as x and y, or as a system of inequalities that apply to the same situation.  As you remember, systems of inequalities must be solved graphically.
 Click here to review solving systems of Linear Inequalities from Math A
(opens in a new window)

How to use your
TI-83+ graphing calculator with systems of  inequalities.
Click calculator.


Applied Problem Using a System of Linear Inequalities
The "We Sell CDs" website plans to purchase ads in a local newspaper advertising their site.  Their operating budget will allow them to spend at most $2200 on this advertising adventure.  They plan to run at most 20 ads.  An ad will cost $50 to appear in the weekday paper and $200 to appear in a weekend edition.  Prepare a graph that will represent all of the possible combinations of ads under these conditions.
Solution:
Let x = the number of weekend ads
Let y = the number of weekday ads

x + y < 20  (there will be at most 20 ads)
200x + 50y < 2200  (the cost of the ads at most $2200)

For this problem x and y cannot be negative numbers, so the answer will be in the first quadrant only.
Solve each of the inequalities above for the y value.
Using your graphing calculator, graph
Y1 = 20 - x  (with icon set to "shade below")
Y2 = (2200-200x)/50  ("shade below" icon set)
Set the window to view only the first quadrant.
Use the intersect option to find the vertices of the quadrilateral whose area makes up the pool of answers (the overlapping shaded region).
 

 


Roberts