6. LINEAR INEQUALITEIS
1. Solve – 12 35 when is a natural number.
Solution: – 12 35 Thus has no value.
2. Solve – 7 12 when is an integer.
Solution: – 7 12 solution set is … , 4, 3, 2.
2.91.
1.71. Thus the
3. Solve 5 26 when is a natural number.
Solution: 5 26 1, 2, 3, 4, 5.
5.2. Thus the solution set is
4. Solve 3 17 when is an integer.
Solution: 3 17 … , 1, 0, 1, 2, 3, 4, 5.
5.67. thus the solution set is
5. Solve 2 3 6 when is an integer.
Solution: 2 3 6 2 6 3 2 3 Thus the solution set is 1, 2, 3, … .
1.5.
6. Solve 3 5 11 when is a real number.
Solution: 3 5 11 3 11 5 3 6 2. Thus
the solution set is 2, ∞.
7. Solve 3 5
Solution: 3 5
"
5 8 when is a natural number.
5 8 5 3 " 5 8 2 " 13
" 6.5. Thus the solution set is 1, 2, 3, 4, 5, 6.
8. Solve 3 5 5 8 when is an integer
Solution: 3 5 5 8 5 3 5 8 2 13
6.5. Thus the solution set is 7, 8, 9, … 9. Solve 3 5
Solution: 3 5
"
5 8 when is a real number.
5 8 5 3 " 5 8 2 " 13
" 6.5. Thus the solution set is ∞, 6.5#.
10. Solve Solution: $
$
$
$
11.
11 $ % $ & $
Thus the solution set is ' , ∞(.
11. Solve
Solution:
$
$
$
"
)
$
"
)
2.
$
2 "
$&*
)
11 7 66 .
4 " 3 8 4 " 3 24
" 24. Thus the solution set is ∞, 24.
12. Solve
Solution:
$ & & $
)
.
)
$ & & $
14 3
203 14 42 60 20 14 20
3. Thus the solution set is +3, ∞.
13. Solve
Solution:
$
'
3( '
3( $
)
)
60 42 34
102
3 5.
3 5 $ & '
( )
3 5
42 15 153 5 8 60 45 75
45 8 75 60
37 15 Thus the solution set is ' , ∞(.
14. Solve 32 5 7
Solution: 32 5 7
6 9
$ & 22 45 3
Solution:
$ & ,$ – & $ % -
$ &
$ &
"
)
9 45
23 3 " 23 "
(.
)
$
"
)
$ & & $ & & $ & $
"
)
"
64 .
$
)
4
11 16 " 45
44 64 " 45 44 45 " 64 99 " 64
99
$
"
$
.
9 5.
9 5 6 15 7
Thus the solution set is '
∞,
15. Solve
)
--
)
. Thus the solution set is ' , ∞(.
--
16. Solve
Solution:
$ & $ & $ & .
.
$ & .
)
$ & –
–
&$
)
$ & , & * % )$
/ /
,
/
&$
$ & $ & .
.
$ & & ) & $
-$ & *
,
/
,
/
202 1 319 18 40 20 57 54
40 57 20 54 17 34 2
2. Thus the solution set ∞, 2.
17. Solve the inequality 2 5 1 5 and represent the solution
graphically on the number line.
Solution: 2 5 1 5 2 5 1 5 7 6 .
The solution set is ' , ∞(. The graphical representation of the solution is
18. Solve the inequality 7 1 " 4 5 and represent the solution
graphically on the number line.
Solution: 7 1 " 4 5 7 4 " 5 1 3 " 4
)
)
" . The solution set is 0'
∞, 01.
19. Mr. A obtained 15 and 18 in first two tests. Find the minimum mark he
should get in the third test to have an average of 16.5.
Solution: Let be the mark he should get in the third test.
By the given condition
2
% $
% * % $
16.5 33 16.5.
49.5 49.5 33 16.5.
Thus the minimum mark he must get is 16.5.
20. Find all pairs of consecutive odd positive integers both of which are lesser
than 18 and such that their sum is more than 15.
Solution: Let and 2 are two consecutive odd positive integers. Now
" 17, 2 " 17. Hence " 15. Also 2 15
2 2 15 2 13 3 6.5.
Thus the possible pairs of consecutive odd positive numbers are
7, 9, 9, 11, 11, 13, 13, 15, 15, 17.
21. The longest side of a triangle is 4 times the shortest side and the third side
is 3 cm shorter than the longest side. If the perimeter of the triangle is atleast
89 cm, find the minimum length of the shortest side.
Solution: Let be the length of the shortest side. Then the longest side is 4.
Also the third side is 4 3. By the given condition, the perimeter
$ % )$ % )$ & 89 9 3
267
9
Thus minimum length of the shortest side is 30 cm.
270 30.
22. A man wants to cut three lengths from a single piece of board of length
115 cm. The second length is to be 5 cm longer than the shortest and the third
length is to be thrice as long as the shortest. Find the possible lengths of the
shortest board if the third piece is to be at least 7 cm longer than the second.
Solution: Let be the length of the shortest piece. Then the other two
lengths are 5 and 3.
By the given condition, 3 5 7 and 5 3 " 113.
Now 3 5 7 2
12 or 6.
Also 5 3 " 113 5 5 " 115 5 " 110
" 21. Thus the length of the shortest piece is
6 cm and " 21 cm.
23. Solve the inequality 2 4 4 and represent the solution region
graphically.
Solution: Consider the straight line 2 4 3 4.
When 3 0, we get, 4 3 4.
When 4 3 0, we get, 3 2.
Thus the line pass through the points 0, 4 and 2, 0.
When 3 0 and 4 3 0 we get, 2 4 3 0 4.
Thus origin lies in the solution region. Thus the solution region is the half plane
lying to the left of the line 2 4 3 4. Also line is not included in the solution
region.
24. Solve the inequality 3 24
6 and represent the solution graphically.
Solution: Consider the straight line 3 24 3 6.
When 3 0, we get, 4 3 3.
When 4 3 0, we get, 3 2.
Thus the line pass through the points 0, 3 and
2, 0.
When 3 0 and 4 3 0 we get, 3 24 3 0 5 6.
Thus origin does not lie in the solution region. Thus the solution region is the
half plane lying to the right of the line 3 24 3 6. Also line is included in the
solution region.
25. Solve the inequality 3
region graphically.
Solution: 3
5 5 and represent the solution
. Thus solution set is 60 , ∞(0.
The solution region is the half plane to the right of the line
3 . Also the line, 3
is included in the solution region.
26. Solve the inequality 24 3 and represent solution set graphically.
Solution: 24 3 4 . Thus the solution set is
'
∞, (. The solution region is the half plane lying below
the line 4 3 .
Also the line 4 3
is not included in the solution region.
27. Solve the inequalities 2 34 12, 2, 4 2 graphically.
Solution: Consider the straight line 2 34 3 12.
When 3 0, 4 3 4 and
when 4 3 0, 3 6.
When 3 0, 4 3 0, we get,
2 34 3 0 12.
Thus solution region is the part of half plane
separated by the line 2 34 3 12 containing
the origin.
The solution region of the set of inequalities is the region bounded by the lines
2 34 3 12, 3 2 and 4 3 2.
28. Solve the inequalities 24 " 6, 4 34 12 graphically.
Solution: Consider the line 24 3 6. This passes through the points 0, 3
and 6, 0. Also when 3 0, 4 3 0, the number 24 3 0 6.
Solution region is the part of the half plane separated by 24 3 6
containing the origin.
Consider the line 4 34 3 12.
This passes through the points 0, 4 and 3, 0.
Also when 3 0, 4 3 0, the number
4 34 3 0 12. Solution region is the part of
the half plane separated by 4 34 3 12 not
containing the origin. The common enclosed region is the solution region. Also
the two lines are included in the solution region.
29. Solve the inequalities 2 4 8, 24 10 graphically.
Solution: Consider the line 2 4 3 8. This passes through the points 0, 8
and 4, 0. Also when 3 0, 4 3 0, the number 2 4 3 0 7 8.
Solution region is the part of the half plane separated by 2 4 3 8 not
containing the origin.
Consider the line 24 3 10. This passes through
the points 0, 5 and 10, 0. Also when
3 0, 4 3 0, the number 24 3 0 7 10.
Solution region is the part of the half plane separated
by 24 3 10 not containing the origin.
The common region of the above two is the solution
region. Also the two lines are not included in the solution region.
30. Solve the inequalities 2 4 " 8, 4 , graphically.
Solution: Consider the line, 2 4 3 8. This passes
through the points 0, 8 and 4, 0. Also when
3 0, 4 3 0, the number 2 4 3 0 8.
Solution region is the part of the half plane separated
by 2 4 3 8 containing the origin.
4 3 is a line passing through the origin. 4 is the
region lying to the left of the line 4 3 . Since, 0 the region lies only in
the first quadrant. The shaded region gives the solution region.
31. Solve the inequalities 2 4
4, 4 " 3, 2 34 " 6
graphically.
Solution: Consider the line, 2 4 3 4. This passes through the points 0, 4
and 2, 0. Also when 3 0, 4 3 0, the number 2 4 3 0 5 4.
Solution region is the part of the half plane separated by 2 4 3 4 not
containing the origin.
Consider the line, 4 3 3. This passes
through the points 0, 3 and 3, 0. Also when
3 0, 4 3 0, the number 4 3 0 " 3.
Solution region is the part of the half plane
separated by 4 3 3 containing the origin.
Consider the line, 2 34 3 6.
This passes through the points 0, 2 and
3, 0. Also when 3 0, 4 3 0, the number 2 34 3 0 " 6.
Solution region is the part of the half plane separated by 2 34 3 6
containing the origin.
The shaded region gives the solution region.
32. Solve the inequalities 24 " 3, 3 44
12, 0, 4
1
graphically.
Solution: Consider the line, 24 3 3. This passes through the points
0, 3⁄2 and 3, 0. Also when 3 0, 4 3 0, the number
24 3 0 " 3. Solution region is the part of the half plane separated by
24 3 3 containing the origin.
Consider the line, 3 44 3 12. This
passes through the points 0, 3 and 4, 0.
Also when 3 0, 4 3 0, the number
3 44 3 0 5 12.
Solution region is the part of the half plane
separated by 3 44 3 12 not containing
the origin. 0 represents the region
lying to the right of the y-axis. 4 3 1 is a
line parallel to x-axis. 4
1 is the region
lying above the line 4 3 1.