◙ EP-Program
5
- Strisuksa School - Roi-et
Math : Real Number and Inequalities
► Dr.Wattana Toutip - Department of Mathematics – Khon Kaen University
© 2010 :Wattana Toutip
◙ wattou@kku.ac.th
◙ http://home.kku.ac.th/wattou
5 Inequalities
An inequality asserts that one expression is less than another .
x  1  3,3x  2 y  4, x 2  3x  7  0 are examples of inequalities.
To solve an inequality is to find the range of values for which it is true. When multiplying or
dividing an inequality by a negative number the inequality changes round .
If x  y then 2 x  2 y .
5.1 Properties of Real Number
5.1.1 Interval and subset of real numbers
 a, b    x a  x  b
 a,     x x  a
 a, b   x a  x  b
[a, )   x x  a
[a, b)   x a  x  b
 ,b    x x  b
(a, b]   x a  x  b
(, b]   x x  b
(, )   x x  R  The set of real numbers
◙ EP .Program – Strisuksa School Roi-et.
Mathematics
5.1.2 Properties of inequalities
Let a, b, c be real numbers
1) Transitive Law:
If a  b and b  c then a  c
If a  b and b  c then a  c
2) Addition Law:
If a  b then a  c  b  c
If a  b then a  c  b  c
3) Multiplicative with Positive Number:
If a  b and c  0 then ac  bc
If a  b and c  0 then ac  bc
4) Multiplicative with Negative Number:
If a  b and c  0 then ac  bc
If a  b and c  0 then ac  bc
5.1.3 Properties of absolute value of real numbers
Let k be a positive real number
1)
x  k if and only if x  k or x  k
2)
x  k if and only if
k  x  k
3)
x  k if and only if
x  k or x  k
5. Real Number and Inequalities
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◙ EP .Program – Strisuksa School Roi-et.
Mathematics
5. Real Number and Inequalities
page 3
5.2 Inequalities in one variable
Linear inequalities are solved by the same algebraic operations as linear equations .Quadratic
and rational inequalities are solved by factorizing.
5.2.1 Example
Solve the following inequalities :
(a) 2 x  3  7
(b) x 2  3x  10  0
1 x
(c)
1
3 x
Solution
(a)
Subtract 3 from both sides, then divide by 2 .
2x  3  3  7  3
2x  4
x2
The solution is: x  2
(b)
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Factorize the quadratic.
( x  5)( x  2)  0
The quadratic is zero at 5 and 2 .These values divide the number line into three regions.
Compile a table to show when the factors are positive and when negative.
x  5
5  x  2
2 x









x5
x2
( x  5)( x  2)
We can see in the real line as shown in the following figure:

5
The solution is: 5  x  2


2
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(c)
Mathematics
5. Real Number and Inequalities
page 4
Take the 1 over to the left hand side .
1 x
1  0
3 x
1  x  (3  x)
0
3 x
2( x  1)
0
3 x
The method of (a) , when expressions are multiplied, applies when they are divide .Proceed
as in (a).
The solution is : 1  x  3
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5.2.2 Exercises
Solve the following inequalities :
1. 3x  1  4
2. 1  2 x  2
3. 1  x  6  2 x
4. 3(1  2 x)  2( x  3)
5. x 2  8 x  12  0
6. x 2  3x  18  0
7. 2 x 2  3x  2  0
8. x 2  x  3  0
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9. x 2  x  0
10. x 2  2 x  35
11. ( x  1)( x  2)( x  3)  0
12. (2 x  1)( x  3)(3  2 x)  0
13. ( x  1)2 ( x  1)  0
14. x 2 (2 x  1)  0
15.
( x  1)( x  2)
0
x3
16.
( x  3)(1  2 x)
0
( x  2)2
17. x  3 
10
x
18.
x 1
1
2 x
19.
x 12

3 x
20.
x  2 x 1

x
x
Mathematics
5. Real Number and Inequalities
page 5
◙ EP .Program – Strisuksa School Roi-et.
Mathematics
5. Real Number and Inequalities
5.3 Absolute Valued with Inequalities
The absolute valued function x is defined by
 x if
x 
 x if
x0
x0
If x  k , for k positive,
Then k  x  k .
If x  k ,for k positive ,
then x  k or x  k .
The graph of y  x is shown in Fig 5.1
Fig 5.1
Inequalities with absolute value terms can be solved by graphs.
5.3.1 Example
Solve the following inequalities
(a) 2 x  1  2
(b) x  3  2 x , by drawing graphs
Solution
(a) Remove the absolute valued signs:
2 x  1  2 or 2 x  1  2
1
1
x   or x  1
2
2
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The solution is: x  
1
1
or x  1
2
2
Mathematics
5. Real Number and Inequalities
page 7
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(b) The graph of y  x  3 and y  2 x are shown. They cross at x  1 .The modulus graph is
below the linear graph after this value.
x 1
The solution is: x  1
5.3.2 Exercises
Solve the following inequalities
1) x  5
2) x  4
3) 2 x  1
4) x  2  4
5) x  3  2
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◙ EP .Program – Strisuksa School Roi-et.
Mathematics
5. Real Number and Inequalities
page 8
Solve questions 6 to 9 by drawing graphs
6) x  x
7) x 
1
x
2
8) x  1  2 x
9) 2 x  x  2
Common errors
1. Solving inequalities
(a) Do not multiply or divide an inequality by term unless you are sure that the is
positive .
(b) If the solution to a quadratic inequality consist of two regions, then leave it like
that .Write x  1or x  3 , do not write 1  x  3 .
2. Inequalities in two dimensions
If you are dealing with say the inequality 3x  2 y  6 , then the region is enclosed by
the line 3x  2 y  6 .It is not enclosed by the lines x  2 and y  3 .
Solution (to exercise)
5.2.2
1. x  1
1
2. x  
2
5
3. x 
3
9
4. x  
4
◙ EP .Program – Strisuksa School Roi-et.
Mathematics
5. Real Number and Inequalities
5. 6  x  2
6. x  3 or x  6
7. all x
8. 2.3  x  1.3
9. 2  x  1 x  5 or x  7
10. x  5 or x  7
11. x  3 or 2  x  1
1
1
12. x  or 1  x  3
2
2
13. x  1
1
14. x   , x  0
2
15. x  3 or 2  x  1
1
16. 3  x  2, 2  x 
2
17. x  5 or 0  x  2
1
18.  x  2
2
19. x  6 or 0  x  6
20. x  0
5.3.2
1. 5  x  5
2. x  4 or x  4
1
1
3.   x 
2
2
4. 2  x  6
5. x  5 or x  1
6. x  0
7. x  0
1
8.   x  1
3
9. x  2 or x  2
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References:
Solomon, R.C. (1997), A Level: Mathematics (4th Edition) , Great Britain, Hillman
Printers(Frome) Ltd.
More:
http://home.kku.ac.th/wattou/service/m456/03.pdf
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