LINEAR INEQUALITIES

Objectives:  To recall how to deal with intervals of real numbers that may be "greater than" or "less than" some fixed real   number.

Recall 1:  An inequality is a comparison of two expressions.  For any two real numbers a and b, there are three possibilities:

1.  a is less than b    (denoted a < b),
2.  a is greater than b    (denoted a > b),
3.  a is equal to b    (denoted a = b).

Recall 2:  The symbol  >= means "greater than or equal to;"
<= means "less than or equal to."

Occassionaly, two inequalities are combined to trap an interval of real numbers.

a < x < b

means that x is greater than a but less than b.

Recall 3:  We may graphically represent intervals of real numbers   that satisfy inequalities; for example,

-3 < x < 5

may be graphically represented by

The parentheses at the ends of the darkened part of the line indicate that  the endpoints -3 and 5 are not included in the region where x satisfies the  inequality -3 < x < 5.

We also use square brackets [ and ] when we wish to  graphically represent the case where the end points are  included; for example, the inequality

0 <= x <= 3

is represented by square brackets because x may be 0 or 3.

Recall 4:  If a < b, and c is any real number, then a + c < b + c.  This   extends to the other kinds of inequalities, <=, >, >=, as well.    This means that

a + c <= b + c,
a + c > b + c,
a + c >= b + c.

This allows us to find the values of x that satisfy inequalities   like

3 < x + 2 <= 5.

These inequalities mean that

x + 2 <= 5  and that 3 < x + 2.

By adding (-2) to both sides of the first inequality, we see that

x + 2 + (-2) <= 5 + (-2)

or that  x <=3

Then, by adding (-2) to the second inequality,

3 + (-2) < x + 2 + (-2),

or  1 < x.

Together this means that 1 < x <= 3.

Recall 5: Multiplying both sides of an inequality by a positive   number produces an equivalent (valid) inequality. Therefore, if a < b, and c is a positive number,

ca < cb.

This fact allows us to find the interval of real numbers x satisfying the  inequalities
-9 < 3x <= 21.

Multiply each term by  1 / 3 to get the equivalent inequalities

(1 / 3)9 < (1 / 3)3x <= (1 / 3)21,

which simplify to

-3 < x <= 7.

Recall 6:  Multiplying both sides of an inequality by a negative number produces an equivalent (valid) inequality, with the  inequality signs reversed.  Therefore, if a < b, and c is a   negative number,

ca > cb.

Recall 7:  We may combine all that we have recalled in this hour to   solve more complicated inequalities.  For example, we are now able to find the interval of x that satisfies

-9 < 3x + 2 <= 21

First add (-2) to both sides to get the equivalent inequality

(-2) + -9 < (-2) + 3x + 2 <= (-2) + 21,

which simplifies to

-11 < 3x <= 19.

Next, multiply both sides by 1 / 3  to get

(1 / 3)(-11) < (1 / 3)(3x) <= (1 / 3)19,

which simplifies to

-11 / 3< x <= 19 / 3.

Examples

Example1. Find the interval of possible values of x in the   inequality

4x - 2 >= 3x + 5

Solution:    Add 2 to both sides of the inequality to get

4x >= 3x + 7.

Next, subtract 3x from both sides to get

x >= 7.

Example 2. Find the interval of possible values of x in the   inequality

-5 < 4x + 1 <= 0.

Solution:  Add 1 to all three parts of the inequality and simplify to   get

-4 < 4x <= 1.

Next, multiply each part by  1 / 4 to get

-1 < x <= 1 / 4 .

Example 3. Find the interval of possible values of x in the   inequality

0 < 5 - 4x <= 3.

Solution:  Add -5 to all three parts of the inequality and simplify to   get

-5 < -4x <= -2.

Next, multiply each part by -1 / 4 to get

5 / 4 > x >= 1 / 2 .

Warning 1:   Always remember to change the direction of the inequality sign after multiplying by a negative number.  Note that if you did not change the direction of the inequality in the above example, you would have ended with the result
5 / 4 < x <= 1/ 2 .
You see that this is absurd because it would imply that  5 / 4 <= 1/ 2 , which in turn would imply that 5 is less than or equal to 2.

Example 4.  Find the interval of possible values of x in the    inequality

(x - 1)(x +3) > 0.

Solution:     We have a product of two terms that is positive.  A product is positive only when both terms are positive or when both terms are negative.  Hence, we are reduced to two cases--

1.        (x - 1) > 0 and (x + 3) > 0
In this case x > 1 and x > -3,
which implies that x > 1.

2.        (x - 1) < 0 and (x + 3) < 0
In this case x < 1 and x < -3,
which implies that x < -3.

We may therefore conclude that x < -3 or x > 1.

Note

The examples that you encountered had  solutions.  Not every inequality that you write will have a  solution.    For example, there are no real numbers x that satisfy the inequality

x^2 < -1

Warning 2:  Your biggest frustrations will occur when you work out an  exercise with haste and carelessness.  It does not take many mistakes to  shake your self confidence; that is a human condition.  Studies show that  most students find math difficult because of a lack of self confidence  rather than a lack of ability.  So take your time and check your results.
It is best to check your results by taking one or two of the values in  your solution to test if they satisfy the original inequality.  In example 4 , you might test whether or not the values x = -4 and x = 2 satisfy  the inequality
(x - 1)(x +3) > 0.

EXERCISES

Find the interval (or intervals) of values of the variable that satisfy the inequalities:

1. x + 5 > 9

2. 4 <= b - 12

3.  x - 3.45 > 2.67

4.  4*y - 8 >= 10

5.  -3x - 5/ 6 < x/3

6.  (2/3)z-5 < 3z

7. (x + 2.3) / 3.2 >= 4.5

8. 4x - 8 >= 7x + 3

9. 3(y - 3) + (y - 3) >= (y - 3)

10. 3(t - 7) < 2 - 10t

11. 0 < 2 - 3x < 5

12.  3(1 - x)(2 + x) > 0

13. (3 - x)(4 + x) < 0

14. 3/8  < (3R - 4)/ 3 <= 3/4

15.  0 < 1 - x/6 <= 1/6

16.  [7(3x - 5) + 3(x - 1)]/3 <= 1 - x/6

17. (x + 4)/ 3 + x/2 > 3(3 - 2x)