Pre-Calculus Algebra
Summary 3
One-dimensional mathematics: inequalities
Interval Notation:
An open interval, denoted by (a, b), consists of all the real numbers for which a < x < b,
that is, all the real number x between a and b.
Example: The interval (-3, 4) includes all the real numbers between -3 and 4 (it does not
include the numbers -3 and 4), that is -3 < x < 4.
-3
4
A closed interval, denoted by [a, b], consists of all the real numbers for which a < x < b,
that is, the numbers a, and b, and all the real number x between a and b.
Example: The interval [-3, 4] includes -3, 4, and all the real numbers between -3 and 4,
that is, -3  x  4.
-3
4
The half-open, or half-closed interval (a, b], consists of all the real numbers for which
a < x < b, that is, all the real numbers between a and b, and the number b.
Example: The interval (-3, 4] includes all the real numbers x between -3 and 4, and the
number 4, that is, -3 < x  4.
-3
4
The half-open, or half-closed interval [a, b), consists of all the real numbers for which
a < x < b, that is, a, and all the real numbers between a and b.
Example: The interval [-3, 4) includes the number -3, and all the real numbers between -3
and 4, that is, -3  x < 4.
-3
4
.
The interval [a,  ), consists of all real numbers for which x  a, that is, the number a,
and all the numbers greater than a.
Example: The interval [-3,  ) includes the number -3, and all the real numbers x greater
than -3, that is x  -3.
-3
.
The interval (a,  ), consists of all real numbers for which x > a, that is, all the numbers
greater than a.
Example: The interval (-3,  ) includes the number -3, and all the real numbers greater
than -3., that is x > -3
.
-3
-1-
The interval (  , b), consists of all real numbers for which x < b, that is, all the numbers
less than b.
Example: The interval (  , 6) includes all the real numbers x less than 6, that is, x < 6.
6
.
The interval (  , b], consists of all real numbers for which x  b, that is, all the numbers
greater than a.
Example: The interval (  , 5] includes all the real numbers less than 5, and the number 5,
that is x  5
5
.
Properties of inequalities:
Properties Addition or subtraction of inequalities:
An inequality remains unchanged if the same number is added or(subtracted) to each side
of the inequality.
Exercises: solve the following inequalities, give the solution set in interval notation.
1. Solve x  3  5
Solution : ( x  3)  3  5  3  x  2
Answer : ( ,  2]
2. Solve the inequality  3  x  2  8
Solution :  3  2  x  2  2  8  2   5  x  6
Answer: (-5, 6]
Properties of multiplication or division by a positive number.
The sense, or direction, of an inequality remains the same if each side is multiplied (or
divided) by the same positive number.
Exercises: solve the following inequalities, give the solution set in interval notation.
3. Solve the inequality 3x  2  7
3x 9
Solution : 3x  2  7  3x  2  2  7  2  3x  9 
 x3
3 3
Answer : ( , 3)
2x  1
4. Solve the inequality  3 
1
4
2x  1
2x  1
Solution :  3 
 1   3 4 
( 4)  1 ( 4)   12  2x  1  4
4
4
 11 2x 5
 11
5
  12  1  2x  1  1  4  1  11  2x  5 



x
2
2 2
2
2
1
1

Answer :   5 , 2 
2
2

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Properties of multiplication and division by a negative number.
The sense, or direction, of an inequality is reversed if each side is multiplied (or divided) by the
same negative number.
Exercises: solve the following inequalities, give the solution set in interval notation.
5. Solve the inequality 3  2x  7
 2x
4
Solution : 3  2x  7  3  2x  3  7  3   2x  4 

 x  2
2
2
Answer : ( 2,  )
2x
1
4
2x
2x
Solution :  3 
 1   3 4 
( 4)  1 ( 4)   12  2  x  4
4
4
 14  1x
2
  12  2  2  x  2  4  2  14   x  2 


 14  x  2
1
1 1
 2  x  14
Answer : [2, 14 )
6. Solve the inequality  3 
Reciprocal property for inequalities:
1
1
If a  0, then  0
and if a  0 then  0
a
a
Exercises: solve the following inequalities, give the solution set in interval notation.
7. Solve the inequality : ( 2x  1) 1  0
1
1
1

 0  2 x  1  0  2x  1  x 
Answer :  ,  
2x  1
2
2

4 3
8. Solve the inequality : 0  
x 2
4 3
4
4 3
4
3
Solution : 0  

 0 and

 x  0 and ( 2x)  ( 2x)
x 2
x
x 2
x
2
8
8
 x  0 and 8  3x  x  0 and 3x  8  x  0 and x 
 x  0 and x 
3
3
8
8

 x
Answer :  ,  
3
3

Solution : ( 2x  1) 1  0 
Solving inequalities with absolute value.
Step 1. Solve the equation.
Step 2. Locate the answers found on step 1 on the number line.
Step 3. By inspection, investigate which intervals of the number line determined by the solutions of
step 1, are solutions of the inequality.
Example 11.
Solve the inequality |x – 3| < 5
Solution:
Step 1. Solve the equation |x – 3| = 5
| x  3 | 5  x  3  5 or x  3  5  x  8 or x  2
-3-
Step 2. Locate the solutions on the number line
-2
8
Step 3.
Investigate the interval ( ,  2) by checking a number like x = -3.
Is |x – 3| < 5?
|-3 -3| = |-6| = 6 is not less than 5. Therefore, the interval ( ,  2) is not a solution.
Investigate the interval ( 2, 8) by checking a number like x = 0.
Is |x – 3| < 5?
|0 -3| = |-3| = 3 is less than 5. Therefore, the interval (-2, 5) is a solution.
Investigate the interval ( 8,  ) by checking a number like x = 10.
Is |x – 3| < 5?
|10 -3| = |7| = 7 is less not than 5. Therefore, the interval ( 8,  ) is not a solution.
Conclusion: the solution of the inequality is the interval (-2, 8)
-2
8
Example 12 Solve the inequality 2x  3  7 .
Solution: 2x  3  7
Step 1. Solve 2x  3  7
| 2x  3 | 7  2x  3  7 or 2x  3  7  2x  10 or 2x  4  x  5 or x  2
Step 2.
-2
5
Step 3. Investigate the interval ( ,  2) by checking a number like x = -3.
Is |2x – 3| > 7?
|-6 -3| = |-9| = 9 is greater than 7. Therefore, the interval ( ,  2] is a solution.
Investigate the interval ( 2, 5) by checking a number like x = 0.
Is |2x – 3| >7?
|0 -3| = |-3| = 3 is not greater than 7. Therefore, the interval (-2, 5) is not a solution.
Investigate the interval ( 5,  ) by checking a number like x = 10.
Is |2x – 3| >7?
|20 -3| = |17| = 17 is greater than 7. Therefore, the interval [5,  ) is a solution.
Conclusion: the solution of the inequality is the interval ( ,  2]  [5,  )
-2
8
-4-
Algebraic and interval notation for inequalities.
Set
1. The set of all real numbers
greater than 5
2. The set of all real numbers
less than 6
3. All real numbers which are
at least 8 (8 or more)
4. All real numbers which are
at most -5
5. All real numbers between -3
and 7
6. All real numbers from -5
to -1
7. All real number from -3 to
5, including 5 but not -3
8. The whole numbers less
than 5
9. The solution of the equation
3x - 2 = 10
10. The solution of the
inequality 3x – 2 >10
11. The solution of the
Algebraic Notation
{x | x>5 }
Interval notation
(5, )
{x | x<6}
(, 6)
x | x  8
[8,  )
x | x  5
(,  5]
x |  3  x  7
(-3, 7)
x |  5  x  1
[-5, -1]
x |  3  x  5
(-3, 5]
{0, 1, 2, 3, 4}
No interval notation
{4}
No interval notation
{x| x >4}
(4, )
{-2, 6}
No interval notation
{-3, 7}
No interval notation
{x |  3  x  7}
(-3, 7)
{x | x  3 or x  7}
(,  3)  (7, )
2
equation x  4 x  12  0
12. The solution of the
equation | x -2| = 5
13. The solution of the
inequality |x -2| < 5
14. The solution of the
inequality |x -2| > 5
-5-