Algebra
Linear Inequations
14
Introduction
You are all familiar with equations in which two sides of a statement (called left hand
side, L.H.S. and right hand side, R.H.S.) are connected by equality sign (=). In fact, you
have solved many equations in one and two variables and also solved some statement (or
word) problems by translating them in the form of equations. A natural question arises:
Is it always possible to translate a statement problem in the form of an equation? Not
always! Instead we may get certain statements involving a sign (or symbol) of inequality
or inequation. These signs are
<, >, ≤, ≥
In this chapter, we shall learn how to solve algebraically linear inequations in one
variable and writing the solution in set notation form. We shall also learn the representation
of solution set on the number line.
4.1 Inequalities among real numbers
Let a, b be any real numbers, then
(1) a is less than b, written as a < b, if and only if b – a is positive. For example,
(i) 2 < 7, since 7 – 2 = 5 which is +ve.
(ii) – 5 < – 2, since – 2 – (– 5) = – 2 + 5 = 3 which is +ve.
(iii)
2 4
4 2 12 − 10
2
which is +ve.
< , since
− =
=
3 5
5 3
15
15
(2) a is less than or equal to b, written as a ≤ b, if and only if b – a is either positive
or zero. For example,
(i) – 3 ≤ 5, since 5 – (– 3) = 5 + 3 = 8 which is +ve.
(ii)
2 2
2 2
= 0.
≤ , since
−
7 7
7 7
(3) a is greater than b, written as a > b, if and only if a – b is positive. For example,
(i) 7 > 2, since 7 – 2 = 5 which is +ve.
(ii) – 2 > – 5, since – 2 – (– 5) = – 2 + 5 = 3 which is +ve.
(iii)
4 2 12 − 10
2
4 2
> , since
which is +ve.
− =
=
5 3
5 3
15
15
Carl Friedrich Gauss
Carl Friedrich Gauss (1777 – 1855), a French mathematician, was a pioneer in the development of
theory of numbers. Mathematics is the queen of all the sciences and number theory is the queen of
mathematics.
(4) a is greater than or equal to b, written as a ≥ b, if and only if a – b is either positive
or zero. For example,
(i) 5 ≥ – 3, since 5 – (– 3) = 5 + 3 = 8 which is +ve.
(ii)
2
2
2 2
≥ , since − = 0.
7
7
7 7
Geometrically, let A, B be the points on the number line representing the numbers
a, b respectively, then
(1) a < b if and only if the point A is towards the left of B.
X′
B
A
•
a
•
b
X
(2) a ≤ b if and only if either the point A is towards the left of B or A coincides
with B.
(3) a > b if and only if the point A is towards the right of B.
X′
A
a•
B
•
b
X
(4) a ≥ b if and only if either the point A is towards the right of B or it coincides
with B.
Note
The statements a < b and a ≤ b are equivalent to the statements b > a and b ≥ a respectively. For
example,
(i) 2 < 7 is equivalent to 7 > 2. (ii) – 3 ≤ 5 is equivalent to 5 ≥ – 3.
4.2 Linear inequations
Statements such as
x < 3, x + 5 ≤ 7, 2x – 3 > 8, 3x + 5 ≥ 11,
x−3
< 2x + 1
2
are called linear inequations in one variable. In general, a linear inequation in one variable
can always be written as
ax + b < 0, ax + b ≤ 0, ax + b > 0 or ax + b ≥ 0
where a and b are real numbers, a ≠ 0.
Replacement set. The set from which values of the variable (involved in the inequation) are
chosen is called the replacement set.
Solution set. A solution to an inequation is a number (chosen from replacement set) which,
when substituted for the variable, makes the inequation true. The set of all solutions of an inequation
is called the solution set of the inequation.
For example, consider the inequation x < 4.
Replacement set
Solution set
(i) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} {1, 2, 3}
(ii) {– 1, 0, 1, 2, 5, 8}
{– 1, 0, 1, 2}
(iii) {– 5, 10}
{– 5}
(iv) {5, 6, 7, 8, 9, 10}
ϕ.
Note that the solution set depends upon the replacement set.
412
Understanding ICSE mathematics – x
4.3 Solving linear inequations in one variable
Solving an inequation in one variable is much like solving an equation in one variable
because most of the basic rules apply. Of course, there is one exception.
4.3.1 Solving linear inequations in one variable
The rules for solving inequations are similar to those for solving linear equations except
for multiplying or dividing by a negative number.
You can do any thing of the following to an inequation:
(i) add (or subtract) the same number or expression to both sides.
(ii)any number (or expression) can be transposed from one side of an inequation to
the other side with the sign of the transposed number (or expression) changed (+ve
to –ve or –ve to +ve).
(iii) multiply (or divide) both sides by the same positive number.
However, when you multiply (or divide) by the same negative number, then the symbol
of inequation is reversed.
For example :
(i) If x < 2, then –x > –2
(Multiplying by –1)
(ii) If 3x – 1 ≥ 5, then – 4 (3x – 1) ≤ – 20
(Multiplying by – 4)
(iii) If – 6x ≤ 12, then x ≥ – 2
(Dividing by – 6)
Thus, always reverse the symbol of an inequation when multiplying or dividing by the same
negative number.
4.3.2 Procedure to solve a linear inequation in one variable
(i) Simplify both sides by removing group symbols and collecting like terms.
(ii)Remove fractions (or decimals) by multiplying both sides by an appropriate factor
(L.C.M. of denominators or a power of 10 in case of decimals).
(iii)Isolate all variable terms on one side and all constants on the other side. Collect
like terms when possible.
(iv)Make the coefficient of the variable 1.
(v) Choose the solution set from the replacement set.
Note
If the replacement set is not given, then it is to be taken as R.
Illustrative Examples
Example 1. Given x ∈ {– 3, – 4, – 5, – 6} and 9 ≤ 1 – 2x, find the possible values of x. Also represent
its solution set on the number line.
Solution. Given 9 ≤ 1 – 2x
⇒ 2x + 9 ≤ 1 – 2x + 2x
(Add 2x)
⇒ 2x + 9 ≤ 1
⇒ 2x + 9 + (– 9) ≤ 1 + (– 9)
(Add – 9)
⇒ 2x ≤ – 8
⇒ x ≤ – 4
(Divide by 2)
But x ∈ {– 3, – 4, – 5, – 6},
∴ the solution set is {– 4, – 5, – 6}.
Linear Inequations
143
The graph of the solution set is shown by thick dots on the number line.
X′
– 6
– 5
– 4
– 3
– 2
– 1
0
1
X
Example 2. Solve the inequation 3 – 2x ≥ x – 12, given that x ∈ N.
Solution. Given 3 – 2x ≥ x – 12
⇒
(– 3) + 3 – 2x ≥ x – 12 + (– 3)
⇒
– 2x ≥ x – 15
⇒
(– x) – 2x ≥ (– x) + x – 15
⇒
– 3x ≥ – 15
⇒
 1
 1  −  (– 3x) ≤  −  (– 15)
3
3
⇒
∴
x≤5
but x ∈ N i.e. x ∈ {1, 2, 3, 4, 5, …},
the solution set is {1, 2, 3, 4, 5}.
[Multiply by – (Add – 3)
(Add – x)
1
and reverse the symbol]
3
Example 3. If the replacement set is the set of integers lying between – 3 and 10, find the solution
set of 14 – 5x ≥ 3x – 40. Also represent the solution set on the number line.
Solution. Given 14 – 5x ≥ 3x – 40
⇒
14 – 5x + 5x ≥ 3x – 40 + 5x
(Add 5x)
14 ≥ 8x – 40
⇒
14 + 40 ≥ 8x – 40 + 40
(Add 40)
⇒
⇒
54 ≥ 8x
8x ≤ 54
( a ≥ b ⇒ b ≤ a)
⇒
⇒
x≤
54
8
(Divide by 8)
⇒
x ≤ 6·75
But the replacement set is the set of integers lying between – 3 and 10
i.e. {– 2, – 1, 0, 1, 2, …, 9},
the solution set is {– 2, – 1, 0, 1, 2, 3, 4, 5, 6}.
∴
The graph of the solution set is shown by thick dots on the number line.
X′
– 3
– 2
– 1
0
1
Example 4. If x ∈ W, find the solution set of
Solution. Given
3
2x − 1
x–
> 1.
5
3
2
3
4
5
6
7
X
3
2x − 1
x–
> 1.
5
3
Multiplying both sides by 15, L.C.M. of 5 and 3, we get
9x – 5(2x – 1) > 15
⇒ 9x – 10x + 5 > 15
⇒ – x > 15 – 5 ⇒ – x > 10
⇒ x < – 10
(Multiplying by – 1)
But x ∈ W,
∴ the solution set is ϕ.
Example 5. Find the values of x, which satisfy the inequation
– 2
414
5 1 2x
< −
≤ 2, x ∈ W. Graph the solution set on the number line.
6 2
3
Understanding ICSE mathematics – x
(2014)
Solution. Given – 2
5 1 2x
17 1 2 x
≤ 2 ⇒ − < −
≤2
< −
6 2
3
6
2
3
⇒
⇒
– 17 < 3 – 4x ≤ 12
– 20 < – 4x ≤ 9
⇒
5 > x ≥ – ⇒
– (Multiplying by 6)
(Adding – 3)
9
4
(Dividing by – 4)
9
≤ x < 5.
4
But x ∈ W i.e. x ∈ {0, 1, 2, 3, 4, …},
∴ the solution set is {0, 1, 2, 3, 4}.
The graph of the solution set is shown by thick dots on the number line.
X′
– 2
–1
0
1
2
3
4
5
6
X
Example 6. Solve the following inequations for real x:
(i)
1 3
1
2x − 1 3x − 2 2 − x
(ii)  x + 4  ≥ (x – 6).
≥
−
 3
25
3
4
5
Solution. (i) Given
2x − 1 3 x − 2 2 − x
.
≥
−
3
4
5
Multiplying both sides by 60, L.C.M. of 3, 4 and 5, we get
20 (2x – 1) ≥ 15 (3x – 2) – 12(2 – x)
⇒40x – 20 ≥ 45 x – 30 – 24 + 12x
⇒40x – 20 ≥ 57x – 54
⇒40x – 57x ≥ – 54 + 20
⇒ – 17x ≥ – 34
⇒ x ≤ 2 As x ∈ R, the solution set = {x : x ∈ R, x ≤ 2}.
(ii) Given
⇒
(Dividing by – 17)
13
 1
 x + 4  ≥ (x – 6)
25
3
x
3
x+2≥
– 2.
3
10
Multiplying both sides by 30, L.C.M. of 10 and 3, we get
9x + 60 ≥ 10x – 60
⇒ 9x – 10x ≥ – 60 – 60
⇒ – x ≥ – 120
⇒ x ≤ 120
As x ∈ R, the solution set = {x : x ∈R, x ≤ 120}.
(Multiplying by – 1)
Example 7. Solve
2x + 1 3x − 2
, x ∈ R. Graph the solution set on the number line.
≥
3
5
Solution. Given
2x + 1 3 x − 2
.
≥
3
5
Multiplying both sides by 15, L.C.M. of 3 and 5, we get
5(2x + 1) ≥ 3(3x – 2) ⇒ 10x + 5 ≥ 9x – 6
⇒
(– 9x) + 10x + 5 ≥ (– 9x) + 9x – 6
x+5≥–6
⇒
⇒
x + 5 + (– 5) ≥ – 6 + (– 5)
⇒
x ≥ – 11.
(Add – 9x)
(Add – 5)
Linear Inequations
145
Hence, the solution set is {x : x ∈ R, x ≥ – 11}.
The graph of the solution set is shown by the thick portion of the number line. The
solid circle at –11 indicates that the number –11 is included among the solutions.
•
0
– 11
X′
X
Example 8. Solve
x 5x − 2 7x − 3
, x ∈R. Also represent the solution set on the number line.
>
−
4
3
5
Solution. Given
x 5x − 2 7 x − 3
.
>
−
4
3
5
Multiplying both sides by 60, L.C.M. of 4, 3 and 5, we get
15x > 20 (5x – 2) – 12 (7x – 3)
⇒ 15x > 100x – 40 – 84x + 36
⇒ 15x > 16x – 4
⇒ 15x – 16x > – 4
⇒ – x > – 4
⇒ x < 4 but x ∈ R.
Hence, the solution set is {x : x ∈R, x < 4}.
The graph of the solution set is shown by thick portion of the number line. The open
circle at 4 indicates that the number 4 is not included among the solutions.
4
0
X′
X
1
3x
2
Example 9. Find the range of values of x, which satisfy the inequation – ≤
+ 1 < , x ∈ R.
5 10
5
Graph the solution set on the number line.
Solution. Given –
1 3x
2
≤
+1< .
5 10
5
Multiplying throughout by 10, L.C.M. of 5, 10 and 5, we get
– 2 ≤ 3x + 10 < 4
⇒ – 2 – 10 ≤ 3x + 10 + (–10) < 4 + (–10)
⇒ –12 ≤ 3x < – 6
⇒ – 4 ≤ x < – 2
Hence, the solution set is {x : x ∈ R, – 4 ≤ x < – 2}.
– 5
X′
– 4
– 3
– 2
– 1
0
(Add – 10)
(Dividing by 3)
X
The graph of the solution set is shown by the thick portion of the number line. The
solid circle at – 4 indicates that the number – 4 is included among the solutions and the open
circle at – 2 indicates that the number – 2 is not included among the solutions.
Example 10. List the elements of the solution set of the inequation – 3 < x – 2 ≤ 9 – 2x, x ∈ N.
Solution. Given – 3 < x – 2 ≤ 9 – 2x, x ∈ N
⇒
– 3 < x – 2 and x – 2 ≤ 9 – 2x
⇒
– 3 + 2 < x – 2 + 2 and 2x + x – 2 ≤ 9 – 2x + 2x
⇒
– 1 < x and 3x – 2 ≤ 9 ⇒ – 1 < x and 3x – 2 + 2 ≤ 9 + 2
⇒
– 1 < x and 3x ≤ 11
⇒
⇒
416
11
– 1 < x and x ≤ 3
11
– 1 < x ≤ 3 but x ∈ N,
∴ the solution set is {1, 2, 3}.
Understanding ICSE mathematics – x
Example 11. P is the solution set of 8x – 1 > 5x + 2 and Q is the solution set of 7x – 2 ≥ 3 (x + 6),
where x ∈ N. Find the set P ∩ Q.
Solution. Given 8x – 1 > 5x + 2
⇒ 8x – 1 + (– 5x + 1) > 5x + 2 + (– 5x + 1)
(Add – 5x + 1)
⇒ 3x > 3 ⇒ x > 1 but x ∈ N,
∴ P = {2, 3, 4, 5, …}.
Also 7x – 2 ≥ 3 (x + 6) ⇒ 7x – 2 ≥ 3x + 18
⇒ 7x – 2 + (– 3x + 2) ≥ 3x + 18 + (– 3x + 2)
(Add – 3x + 2)
⇒ 4x ≥ 20 ⇒ x ≥ 5 but x ∈ N,
∴Q = {5, 6, 7, 8, …}.
∴ P ∩ Q = {5, 6, 7, 8, …}.
Example 12. Solve the following inequation, and graph the solution on the number line :
2x – 5 ≤ 5x + 4 < 11, x ∈ R.
Solution. Given 2x – 5 ≤ 5x + 4 < 11, x ∈ R
⇒ 2x – 5 ≤ 5x + 4 and 5x + 4 < 11
⇒ 2x – 5 + (– 5x + 5) ≤ 5x + 4 + (– 5x + 5) and 5x + 4 + (– 4) < 11 + (– 4)
⇒ – 3x ≤ 9 and 5x < 7
(2002)
7
⇒ x ≥ – 3 and x < 5
7
⇒ – 3 ≤ x and x < 5
7
⇒ – 3 ≤ x < 5
where x ∈ R.
7
∴ The solution set =  x : x ∈ R , − 3 ≤ x <  .
5

The graph of the solution set is shown by the thick portion of the number line. The
solid circle at – 3 indicates that the number – 3 is included among the solutions and the open
7
7
circle at 5 indicates that the number 5 is not included among the solutions.
X′
– 3
– 2
– 1
0
1
7
5
X
Example 13. Solve the following inequation and write the solution set:
13x – 5 < 15x + 4 < 7x + 12, x ∈ R.
Represent the solution on a real number line.
Solution. Given 13x – 5 < 15x + 4 < 7x + 12, x ∈ R
⇒ 13x – 5 < 15x + 4 and 15x + 4 < 7x + 12
⇒ 13x – 15x < 4 + 5 and 15x – 7x < 12 – 4
⇒ – 2x < 9 and 8x < 8
⇒ x > –
(2015)
9
and x < 1
2
⇒ –
9
< x and x < 1
2
⇒ –
9
< x < 1, x ∈ R.
2
Hence, the solution set = {x : x ∈ R, –
9
< x < 1}
2
Linear Inequations
147
– 9
2
– 5
X′
– 4
– 3
– 2
– 1
0
1
2
3
X
The graph of the solution set is shown by the thick portion of the number line. The open
circles at –
9
and at 1 indicate that these numbers are not included in the solution set.
2
Example 14. Solve the following inequation, write the solution set and represent it on the number
x x
1 1
≤ − 1 < , x ∈ R.
3 2
3 6
1 1
x x
Solution. Given − ≤ − 1 < , x ∈ R
3 2
3 6
(2013)
line: −
⇒ −
x x 4 1
⇒ – 2x ≤ 3x – 8 < 1
≤ − <
3 2 3 6
(Multiplying by 6)
⇒ – 2x ≤ 3x – 8 and 3x – 8 < 1
⇒ 8 ≤ 5x and 3x < 9
⇒
8
8
≤ x and x < 3 ⇒
≤ x < 3, x ∈ R
5
5
8
∴ The solution set =  x : x ∈ R , ≤ x < 3  .
5


The graph of the solution set is shown by the thick portion of the number line.
8
5
– 2
X′
– 1
0
1
2
3
4
X
5
Example 15. Find the set of values of x satisfying
7x + 3 ≥ 3x – 5 and
x
5
– 5 ≤ – x, where x ∈ N.
4
4
Also graph the solution set on the number line.
Solution. Given 7x + 3 ≥ 3x – 5 and
x
5
– 5 ≤ – x
4
4
⇒ 7x – 3x ≥ – 5 – 3 and x – 20 ≤ 5 – 4x
⇒ 4x ≥ – 8 and x + 4x ≤ 5 + 20
⇒ x ≥ – 2 and 5x ≤ 25
⇒ – 2 ≤ x and x ≤ 5
⇒ – 2 ≤ x ≤ 5.
But x ∈ N i.e. x ∈ {1, 2, 3, 4, …},
∴ the solution set is {1, 2, 3, 4, 5}.
The graph of the solution set is shown by thick dots on the number line.
X′
– 1
0
1
2
3
4
5
6
X
Example 16. John needs a minimum of 360 marks in four tests in a Mathematics course to obtain
an A grade. On his first three tests, he scored 88, 96, 79 marks. What should his score be in the fourth
test so that he can make an A grade?
Solution. Let John score x marks in the fourth test. Then the sum of John’s test scores
should be greater than or equal to 360 i.e.
418
88 + 96 + 79 + x ≥ 360
⇒ x ≥ 360 + (– 263)
⇒
263 + x ≥ 360
⇒
x ≥ 97.
Understanding ICSE mathematics – x
∴ John should score 97 marks or greater than 97 marks in the fourth test to obtain A
grade.
Example 17. An integer is such that one-third of the next integer is atleast 2 more than one-fourth
of the previous integer. Find the smallest value of the integer.
Solution. Let the integer be x, then one-third of the next integer is
of the previous integer is
According to given,
⇒
⇒
⇒
⇒
∴
x−1
.
4
x+1
and one-fourth
3
x+1 x−1
+2
≥
3
4
4 (x + 1) ≥ 3(x – 1) + 2 × 12
4x + 4 ≥ 3x – 3 + 24
4x – 3x ≥ – 3 + 24 – 4
x ≥ 17.
The smallest value of x = 17.
Example 18. Find three largest consecutive natural numbers such that the sum of one-third of first,
one-fourth of second and one-fifth of the third is atmost 25.
Solution. Let three required consecutive natural numbers be x, x + 1 and x + 2.
According to given,
x x+1 x+2
≤ 25
+
+
3
4
5
⇒ 20x + 15 (x + 1) + 12 (x + 2) ≤ 25 × 60
⇒ 20x + 15x + 15 + 12x + 24 ≤ 1500
⇒ 47x + 39 ≤ 1500
⇒ 47x ≤ 1500 – 39
⇒ 47x ≤ 1461 ⇒ x ≤
⇒ x ≤ 31
4
.
47
1461
47
∴ The largest value of x = 31.
When x = 31, x + 1 = 31 + 1 = 32 and x + 2 = 31 + 2 = 33.
Hence, the required natural numbers are 31, 32 and 33.
Exercise 4
1. Solve the inequation 3x – 11 < 3, where x ∈ {1, 2, 3, …, 10}. Also represent its solution on
a number line.
2. Solve 2(x – 3) < 1, x ∈ {1, 2, 3, …, 10}.
3. Solve 5 – 4x > 2 – 3x, x ∈ W. Also represent its solution on the number line.
4. List the solution set of 30 – 4(2x – 1) < 30, given that x is a positive integer.
5. Solve 2(x – 2) < 3x – 2, x ∈ {– 3, – 2, – 1, 0, 1, 2, 3}.
6. If x is a negative integer, find the solution set of
7. Solve
8. Solve x – 3 (2 + x) > 2(3x – 1), x ∈ {– 3, – 2, – 1, 0, 1, 2}. Also represent its solution on the
number line.
9. Given x ∈ {1, 2, 3, 4, 5, 6, 7, 9}, solve x – 3 < 2x – 1.
2 1
+ (x + 1) > 0.
3 3
2x − 3 1
≥ , x ∈ {0, 1, 2, …, 8}.
4
2
Linear Inequations
149