Earlier we have learnt to construct linear equations. Let us now study about ‘Inequations’.

**Inequations**

Earlier we have learnt to construct linear
equations. Let us now study about ‘Inequations’.

As per the norms the minimum age limit
to own a driving licence is 18 years.

Therefore, when Rajiv owns a driving
licence, we can say that he is *at least* 18 years of age.

Now, if Rajiv’s age is represented by
*x*, then this statement can be written mathematically as *x* ≥ 18, that
is, we are not sure about his age, still we can say that his age is greater than
or equal to 18 years.

Similarly, the statement *‘This jug
can hold up to 5 litres of water’* can be written mathematically as *x*
≤ 5, where ‘*x’* represents the volume of water in the jug.

We know that, the sum of the measures
of any two sides of a triangle is greater than the measure of its third side. Thus,
if the measure of the three sides are represented by *a*,* b *and* c
**units*, then we may write this fact
as* a *+* b *>* c*,* a *+* c *>* b *and* b
*+* c *>* a*.

When *x* ≠ 10, then either *x*
> 10 or *x* < 10. That is, if the value of the variable ‘*x’* is
not 10, then the value of ‘*x’* will either be greater than 10 or be less than
10.

An algebraic statement that shows two
algebraic expressions being unequal is known as an algebraic inequation.

In general, when two expressions are
compared, one might be; less than (<), less than or equal to (≤), greater than
(>), greater than or equal to (≥) the other.

In an inequation, the algebraic expressions
are connected by one out of the four signs of inequalities, namely, >, ≥, <
and ≤.

**Try these**

Construct inequations for
the following statements:

1. Ramesh’s salary is more than ₹25,000 per month.

2. This lift can carry maximum of 5 persons.

3. The exhibition will
be there in town for at least 100 days.

**Solution: **

**1.** x > 25,000, where *x*
is Ramesh’s Salary per month.

**2.** y ≤ 5, where *y* is the
maximum number of persons the left can carry.

3. z ≥ 100, where *z* is
the number of days when the exhibition is there.

** **

__1. Solving linear Inequations__

A simple linear equation has atmost one
solution, but a linear inequation may have many solutions.

To solve an inequation, it is necessary
to know the set of values that the variable symbol can be substituted with. The
collection of of all such values of an inequation is known as **solution**
of the inequation.

For example, the solution of the equation
3*x* – 3= 12 is 5.
(How?) Let us find the solution for the inequation 3*x* − 3 < 12, where
*x* is a natural number. Note that, the solutions of this inequation are ‘natural
numbers’. Now,

Add 3 on both sides, we get 3*x*
− 3 + 3 < 12 + 3⇒
3*x* < 15

Divide by 3 on both sides, we get 3*x*/3 < 15/3 ⇒
*x*<5

Hence, *x* takes value which is
less than 5 and *x* is a natural
number. Thus, the solution for this inequation are 1, 2, 3 and 4.

**Note**

When ‘*x*’ is not
restricted to a natural number, the solution includes all values less than 5.

**Rules to solve inequation**

While solving an inequation, the rules
for transposition in case of inequalities are the same as for equations.

1**. Addition** of the same number on both sides
of the inequation does not change the
value of the inequation. Example: 10 > 5 ⇒ 10 + 1 > 5 + 1 ⇒
11 > 6.

Extending this result, when adding any
number ‘*x*’ instead of 1, the inequality
10 +
*x* > 5 + *x* remains unchanged.

**2.
Subtraction** of the same
number from both sides of the inequation does not change the value of the inequation.
Example: 10 > 5 ⇒ 10−1>5−1
⇒
9>4. * *

Extending this result, when subtracting
any number *x *instead of 1, the
inequatity 10−*x *> 5−*x* remains unchanged.

3. **Multiplication **by the same**
***positive number*** **on both sides of the inequation does not change
the value of the inequation. Example: 10 > 5 ⇒10×**2** > 5×**2** ⇒ 20 > 10.

Similarly, when multiplying any positive number *x* instead
of 2, the inequatity 10 × *x* > 5×*x* remains unchanged.

**4.
Division** by the same *non-zero* *positive number* on both sides of the inequation does not change the
value of the inequation. Example: 10 > 5 ⇒ 10/5 > 5/5 ⇒ 2 > 1.

Similarly, when dividing any non zero
positive number *x* instead of 5, the inequatity 10/*x* > 5/*x
*remains unchanged.

**Note**

When both sides of an inequation
are multiplied or divided by the same *non-zero* *negative number*, the
sign of inequality is reversed. For example, consider 3 < 12.

Multiplying −1 on both
sides, we get, 3 × (−1) < 12 × (−1)

−3 < −12.

But, it is not true. Because,
−3 is greater than −12.

So, −3 **>** −12.
Note that the sign of inequality is reversed.

To generalize , when *x*
< *y* is multiplied by –1 on both sides , we
get −*x* > −*y* .

Interchanging the expressions
on both sides of an equation, does not make any change in the equation. For example,
*x* + 3 =5 and 5 = *x* +3 both are same.

But, if the expressions
on both sides of an inequation are interchanged, the sign of inequality must be
reversed.

For example, 30 > 20
is the same as 20 < 30 and −18 < −9 is the same as −9 > −18.

__Example 3.10 __

Solve: 2*x*** **+ 4 < 18, where** ***x*** **is a natural number.

**Solution:**

2*x* + 4 < 18

2*x +* 4 – 4 < 18 − 4 [Subtracting 4 from both sides]

2*x* < 14 [Divide by 2 on both sides]

*x*< 7

Since the solution belongs to natural
numbers, that are less than 7, we take the values of the *x* as 1, 2, 3, 4,
5 and 6.

Therefore, the solutions are 1, 2, 3,
4, 5 and 6.

__Example 3.11__

** **Solve: 5 − 7*x*** **≥ 33, where** ***x*** **is an integer.

**Solution:**

5 − 7*x* ≥ 33

5 − 5 − 7*x* ≥ 33 − 5 [Subtracting 5 from both sides]

−7*x* ≥ 28

−7x / −7 ≥ 28 / −7
[Dividing both sides by –7]

*x *≤ −4* *[since,
it is divided by a negative number, the inequality is reversed]

Since, solution belongs to the set of
integers, that are less than −4, we take the values of *x* as –4, –5, –6, ...

Therefore, the solutions are –4, –5,
–6,... .

**Think**

Hameed saw a stranger in
the street. He told his parent, “The stranger’s age is between 40 to 45 years, and
his height is between 160 to 170 *cm*”.

Convert the above verbal
statement into an algebraic inequations by using *x* and *y* as variables
of age and height.

**Solution: **

Let *x* be the age and *y* be the height then

40 ≤ *x* ≤45 and 160 ≤ *y* ≤ 170

__Example 3.12__

If one worker earns ₹200
per day, how many workers can be employed** **within a monthly budget of ₹3 Lakh?

**Solution:**

Let the number of workers be *x*.

Then, the wages that* x *workers will earn per day = ₹200x

The wages that* x *workers will earn per month= ₹(200*x* × 30) = ₹6000*x*

Given that, this amount cannot exceed
₹300000.

Otherwise, it can be written as 6000*x*
≤ 300000

6000/6000*x* ≤
300000/6000 [Divide by 6000 on both sides]

* x *≤ 50

Thus, up to 50 workers can be employed
on a monthly budget of ₹300000.

__2. Graphical representation of Inequation__

The solutions of an inequation can be
represented on the number line by marking the true values of solutions with different
colour on the number line.

Look into the following inequations and
its graphical representation on number line. Here, we consider the solution belongs
to natural numbers. That is, each and every value of the solution is a natural number.

1. When *x* < 3, the solution
in natural numbers are 1 and 2. Its graph on number line is shown below:

2. When *x* ≥ 3, the solutions are
natural numbers 3, 4, 5, ... and its graph is as shown below:

3. To mark the values represented by
the inequation 2 ≤ *x* ≤ 5, the solutions are set of natural numbers 2, 3,
4 and 5 and its graph is as given below:

__Example 3.13__

Represent the solutions −8 < 2*x*
< 10 in a number line, where *x* is a natural number.

**Solution:**

−8 < 2*x* < 10

−8/2 < 2*x*/2 < 10/2 [Dividing the inequation by
2]

−4 <*
x *< 5

Since the solution belongs to the set
of natural numbers, the solutions are 1, 2, 3 and 4. It’s graph on the number line
is shown below:

**Note**

Since the solution is restricted
to natural numbers, −3, −2, −1 and 0 have not been marked as solutions.

__Example 3.14 __

Represent the solutions of
3*x*** **+ 9 ≤ 12 in a number line,
where** ***x*** **is an** **integer.

**Solution:**

3x + 9 ≤ 12

3*x*/3
+ 9/3 ≤ 12/3 [Dividing the inequation by 3 on both sides]

*x
*+ 3 ≤ 4

x + 3 – 3 ≤ 4–3 [Subtracting 3 from both sides]

x ≤ 1

Since the solution belongs to integers,
the solutions are 1, 0, −1, −2, …. It’s graph on the number line is shown below:

__Example 3.15 __

Solve the inequation: −2
≤** ***z*** **+ 3 ≤ 5, where** ***z*** **is an integer. Also, represent** **the solution, graphically.

**Solution:**

−2 ≤ *z* + 3
≤ 5

−2 −3 ≤ *z* + 3 −3 ≤ 5−3 [Subtracting −3 from the
inequation]

−5 ≤ z ≤ 2

Since the solution belongs to integers,
the solutions are –5, –4 , –3, –2, –1, 0, 1 and 2. It’s graph on the number line
is shown below:

__Example 3.16__

** **Solve graphically: 6*y*** **− 5 ≤ 2*y*** **+ 7, where** ***y*** **is an integer.

**Solution:**

6y − 5 ≤ 2y + 7

6y − 2y − 5 ≤ 2y − 2y + 7 [Subtracting 2y from both sides]

4y − 5 ≤ 7

4y − 5 + 5 ≤ 7+5 [Adding 5 on both sides]

4y ≤ 12

4y/4 ≤ 12/4 [Dividing
by 4 on both sides]

y ≤ 3

Since the solution belongs to integers,
the solutions are 3, 2, 1, 0, –1, –2,..... It’s graph on the number line is shown
below:

Tags : Algebra | Term 3 Chapter 3 | 7th Maths , 7th Maths : Term 3 Unit 3 : Algebra

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