NCERT Textbook- Linear Equations in One Variable Class 8 Notes | EduRev

 Page 1


LINEAR EQUATIONS IN ONE VARIABLE  21
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
537
2,6102
22
yz += + =-
Y ou would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2xy + 5
has two variables. W e however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
Linear Equations in
One Variable
CHAPTER
2
2x – 3 = 7
2x – 3 = LHS
7 = RHS
Page 2


LINEAR EQUATIONS IN ONE VARIABLE  21
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
537
2,6102
22
yz += + =-
Y ou would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2xy + 5
has two variables. W e however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
Linear Equations in
One Variable
CHAPTER
2
2x – 3 = 7
2x – 3 = LHS
7 = RHS
22  MATHEMATICS
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
W e assume that the two sides of the equation are balanced.
W e perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations which have Linear Expressions
on one Side and Numbers on the other Side
Let us recall the technique of solving equations with some examples. Observe the solutions;
they can be any rational number.
Example 1: Find the solution of 2x – 3 = 7
Solution:
Step 1 Add 3 to both sides.
2x – 3 + 3 = 7 + 3 (The balance is not disturbed)
or 2x =10
Step 2 Next divide both sides by 2.
2
2
x
 =
10
2
or x = 5 (required solution)
Example 2: Solve 2y + 9 = 4
Solution: Transposing 9 to RHS
2y  = 4 – 9
or 2y  =– 5
Dividing both sides by 2, y =
5
2
-
(solution)
To check the answer: LHS = 2 
5
2
- ??
??
??
 + 9  = – 5 + 9 = 4 = RHS (as required)
Do you notice that the solution 
5
2
- ??
??
??
 is a rational number? In Class VII, the equations
we solved did not have such solutions.
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
Page 3


LINEAR EQUATIONS IN ONE VARIABLE  21
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
537
2,6102
22
yz += + =-
Y ou would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2xy + 5
has two variables. W e however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
Linear Equations in
One Variable
CHAPTER
2
2x – 3 = 7
2x – 3 = LHS
7 = RHS
22  MATHEMATICS
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
W e assume that the two sides of the equation are balanced.
W e perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations which have Linear Expressions
on one Side and Numbers on the other Side
Let us recall the technique of solving equations with some examples. Observe the solutions;
they can be any rational number.
Example 1: Find the solution of 2x – 3 = 7
Solution:
Step 1 Add 3 to both sides.
2x – 3 + 3 = 7 + 3 (The balance is not disturbed)
or 2x =10
Step 2 Next divide both sides by 2.
2
2
x
 =
10
2
or x = 5 (required solution)
Example 2: Solve 2y + 9 = 4
Solution: Transposing 9 to RHS
2y  = 4 – 9
or 2y  =– 5
Dividing both sides by 2, y =
5
2
-
(solution)
To check the answer: LHS = 2 
5
2
- ??
??
??
 + 9  = – 5 + 9 = 4 = RHS (as required)
Do you notice that the solution 
5
2
- ??
??
??
 is a rational number? In Class VII, the equations
we solved did not have such solutions.
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
LINEAR EQUATIONS IN ONE VARIABLE  23
Example 3: Solve 
5
32
x
+
 = 
3
2
-
Solution: Transposing 
5
2
to the RHS, we get 
3
x
 = 
35 8
22 2
-
-=-
or
3
x
 =– 4
Multiply both sides by 3, x = – 4 × 3
or x = – 12 (solution)
Check: LHS = 
12 5 5 8 5 3
4
32 2 2 2
-+ -
-+ =- + = = =
 RHS (as required)
Do you now see that the coefficient of a variable in an equation need not be an integer?
Example 4: Solve 
15
4
 – 7x = 9
Solution: We have
15
4
 – 7x =9
or – 7x = 9 – 
15
4
(transposing 
15
4
 to R H S)
or – 7x =
21
4
or x =
21
4( 7) ×-
(dividing both sides by – 7)
or x =
37
47
×
-
×
or x =
3
4
-
(solution)
Check:  LHS = 
15 3
7
44
- ??
-
??
??
= 
15 21 36
9
44 4
+= =
 =  RHS (as required)
EXERCISE 2.1
Solve the following equations.
1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2
4.
317
77
x +=
5. 6x = 12 6.
10
5
t
=
7.
2
18
3
x
=
8. 1.6 = 
1.5
y
9. 7x – 9 = 16
Page 4


LINEAR EQUATIONS IN ONE VARIABLE  21
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
537
2,6102
22
yz += + =-
Y ou would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2xy + 5
has two variables. W e however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
Linear Equations in
One Variable
CHAPTER
2
2x – 3 = 7
2x – 3 = LHS
7 = RHS
22  MATHEMATICS
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
W e assume that the two sides of the equation are balanced.
W e perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations which have Linear Expressions
on one Side and Numbers on the other Side
Let us recall the technique of solving equations with some examples. Observe the solutions;
they can be any rational number.
Example 1: Find the solution of 2x – 3 = 7
Solution:
Step 1 Add 3 to both sides.
2x – 3 + 3 = 7 + 3 (The balance is not disturbed)
or 2x =10
Step 2 Next divide both sides by 2.
2
2
x
 =
10
2
or x = 5 (required solution)
Example 2: Solve 2y + 9 = 4
Solution: Transposing 9 to RHS
2y  = 4 – 9
or 2y  =– 5
Dividing both sides by 2, y =
5
2
-
(solution)
To check the answer: LHS = 2 
5
2
- ??
??
??
 + 9  = – 5 + 9 = 4 = RHS (as required)
Do you notice that the solution 
5
2
- ??
??
??
 is a rational number? In Class VII, the equations
we solved did not have such solutions.
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
LINEAR EQUATIONS IN ONE VARIABLE  23
Example 3: Solve 
5
32
x
+
 = 
3
2
-
Solution: Transposing 
5
2
to the RHS, we get 
3
x
 = 
35 8
22 2
-
-=-
or
3
x
 =– 4
Multiply both sides by 3, x = – 4 × 3
or x = – 12 (solution)
Check: LHS = 
12 5 5 8 5 3
4
32 2 2 2
-+ -
-+ =- + = = =
 RHS (as required)
Do you now see that the coefficient of a variable in an equation need not be an integer?
Example 4: Solve 
15
4
 – 7x = 9
Solution: We have
15
4
 – 7x =9
or – 7x = 9 – 
15
4
(transposing 
15
4
 to R H S)
or – 7x =
21
4
or x =
21
4( 7) ×-
(dividing both sides by – 7)
or x =
37
47
×
-
×
or x =
3
4
-
(solution)
Check:  LHS = 
15 3
7
44
- ??
-
??
??
= 
15 21 36
9
44 4
+= =
 =  RHS (as required)
EXERCISE 2.1
Solve the following equations.
1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2
4.
317
77
x +=
5. 6x = 12 6.
10
5
t
=
7.
2
18
3
x
=
8. 1.6 = 
1.5
y
9. 7x – 9 = 16
24  MATHEMATICS
10. 14y – 8 = 13 11. 17 + 6p = 9 12.
7
1
315
x
+=
2.3 Some Applications
W e begin with a simple example.
Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the
numbers?
We have a puzzle here. We do not know either of the two numbers, and we have to
find them. W e are given two conditions.
(i) One of the numbers is 10 more than the other.
(ii) Their sum is 74.
W e already know from Class VII how to proceed. If the smaller number is taken to
be x, the larger number is 10 more than x, i.e., x + 10. The other condition says that
the sum of these two numbers x and x + 10 is 74.
This means that x + (x + 10) = 74.
or 2x + 10 = 74
Transposing 10 to RHS, 2x = 74 – 10
or 2x =64
Dividing both sides by 2, x = 32.  This is one number.
The other number is x + 10 = 32 + 10 = 42
The desired numbers are 32 and 42.  (Their sum is indeed 74 as given and also one
number is 10 more than the other.)
W e shall now consider several examples to show how useful this method is.
Example 5: What should be added to twice the rational number 
7
3
-
 to get 
3
7
?
Solution:  T wice the rational number 
7
3
-
 is 
714
2
33
-- ??
×=
??
??
. Suppose x added to this
number gives 
3
7
; i.e.,
14
3
x
- ??
+
??
??
 =
3
7
or
14
3
x -
 =
3
7
or x =
314
73
+
(transposing 
14
3
 to RHS)
=
(3 3) (14 7)
21
×+ ×
 = 
9 98 107
21 21
+
=
.
Page 5


LINEAR EQUATIONS IN ONE VARIABLE  21
2.1  Introduction
In the earlier classes, you have come across several algebraic expressions and equations.
Some examples of expressions we have so far worked with are:
5x, 2x – 3, 3x + y, 2xy + 5, xyz + x + y + z, x
2
 + 1, y + y
2
Some examples of equations are: 5x = 25, 2x – 3 = 9, 
537
2,6102
22
yz += + =-
Y ou would remember that equations use the equality (=) sign; it is missing in expressions.
Of these given expressions, many have more than one variable. For example, 2xy + 5
has two variables. W e however, restrict to expressions with only one variable when we
form equations. Moreover, the expressions we use to form equations are linear . This means
that the highest power of the variable appearing in the expression is 1.
These are linear expressions:
2x, 2x + 1, 3y – 7, 12 – 5z, 
5
( – 4) 10
4
x +
These are not linear expressions:
x
2
 + 1, y + y
2
, 1 + z + z
2
 + z
3
(since highest power of variable > 1)
Here we will deal with equations with linear expressions in one variable only. Such
equations are known as linear equations in one variable. The simple equations which
you studied in the earlier classes were all of this type.
Let us briefly revise what we know:
(a) An algebraic equation is an equality
involving variables. It  has an equality sign.
The expression on the left of the equality sign
is the Left Hand Side (LHS). The expression
on the right of the equality sign is the Right
Hand Side (RHS).
Linear Equations in
One Variable
CHAPTER
2
2x – 3 = 7
2x – 3 = LHS
7 = RHS
22  MATHEMATICS
(b) In an equation the values of
the expressions on the LHS
and RHS are equal. This
happens to be true only for
certain values of the variable.
These values are the
solutions of the equation.
(c) How to find the solution of an equation?
W e assume that the two sides of the equation are balanced.
W e perform the same mathematical operations on both
sides of the equation, so that the balance is not disturbed.
A few such steps give the solution.
2.2 Solving Equations which have Linear Expressions
on one Side and Numbers on the other Side
Let us recall the technique of solving equations with some examples. Observe the solutions;
they can be any rational number.
Example 1: Find the solution of 2x – 3 = 7
Solution:
Step 1 Add 3 to both sides.
2x – 3 + 3 = 7 + 3 (The balance is not disturbed)
or 2x =10
Step 2 Next divide both sides by 2.
2
2
x
 =
10
2
or x = 5 (required solution)
Example 2: Solve 2y + 9 = 4
Solution: Transposing 9 to RHS
2y  = 4 – 9
or 2y  =– 5
Dividing both sides by 2, y =
5
2
-
(solution)
To check the answer: LHS = 2 
5
2
- ??
??
??
 + 9  = – 5 + 9 = 4 = RHS (as required)
Do you notice that the solution 
5
2
- ??
??
??
 is a rational number? In Class VII, the equations
we solved did not have such solutions.
x = 5 is the solution of the equation
2x – 3 = 7. For  x = 5,
LHS = 2 × 5 – 3 = 7 = RHS
On the other hand x = 10 is not a solution of the
equation. For x = 10,  LHS = 2 × 10 –3 = 17.
This is not equal to the RHS
LINEAR EQUATIONS IN ONE VARIABLE  23
Example 3: Solve 
5
32
x
+
 = 
3
2
-
Solution: Transposing 
5
2
to the RHS, we get 
3
x
 = 
35 8
22 2
-
-=-
or
3
x
 =– 4
Multiply both sides by 3, x = – 4 × 3
or x = – 12 (solution)
Check: LHS = 
12 5 5 8 5 3
4
32 2 2 2
-+ -
-+ =- + = = =
 RHS (as required)
Do you now see that the coefficient of a variable in an equation need not be an integer?
Example 4: Solve 
15
4
 – 7x = 9
Solution: We have
15
4
 – 7x =9
or – 7x = 9 – 
15
4
(transposing 
15
4
 to R H S)
or – 7x =
21
4
or x =
21
4( 7) ×-
(dividing both sides by – 7)
or x =
37
47
×
-
×
or x =
3
4
-
(solution)
Check:  LHS = 
15 3
7
44
- ??
-
??
??
= 
15 21 36
9
44 4
+= =
 =  RHS (as required)
EXERCISE 2.1
Solve the following equations.
1. x – 2 = 7 2. y + 3 = 10 3. 6 = z + 2
4.
317
77
x +=
5. 6x = 12 6.
10
5
t
=
7.
2
18
3
x
=
8. 1.6 = 
1.5
y
9. 7x – 9 = 16
24  MATHEMATICS
10. 14y – 8 = 13 11. 17 + 6p = 9 12.
7
1
315
x
+=
2.3 Some Applications
W e begin with a simple example.
Sum of two numbers is 74. One of the numbers is 10 more than the other. What are the
numbers?
We have a puzzle here. We do not know either of the two numbers, and we have to
find them. W e are given two conditions.
(i) One of the numbers is 10 more than the other.
(ii) Their sum is 74.
W e already know from Class VII how to proceed. If the smaller number is taken to
be x, the larger number is 10 more than x, i.e., x + 10. The other condition says that
the sum of these two numbers x and x + 10 is 74.
This means that x + (x + 10) = 74.
or 2x + 10 = 74
Transposing 10 to RHS, 2x = 74 – 10
or 2x =64
Dividing both sides by 2, x = 32.  This is one number.
The other number is x + 10 = 32 + 10 = 42
The desired numbers are 32 and 42.  (Their sum is indeed 74 as given and also one
number is 10 more than the other.)
W e shall now consider several examples to show how useful this method is.
Example 5: What should be added to twice the rational number 
7
3
-
 to get 
3
7
?
Solution:  T wice the rational number 
7
3
-
 is 
714
2
33
-- ??
×=
??
??
. Suppose x added to this
number gives 
3
7
; i.e.,
14
3
x
- ??
+
??
??
 =
3
7
or
14
3
x -
 =
3
7
or x =
314
73
+
(transposing 
14
3
 to RHS)
=
(3 3) (14 7)
21
×+ ×
 = 
9 98 107
21 21
+
=
.
LINEAR EQUATIONS IN ONE VARIABLE  25
Thus 
107
21
 should be added to 
7
2
3
- ??
×
??
??
 to give 
3
7
.
Example 6: The perimeter of a rectangle is 13 cm and its width is 
3
2
4
 cm. Find its
length.
Solution:  Assume the length of the rectangle to be x cm.
The perimeter of the rectangle = 2 × (length + width)
=2 × (x + 
3
2
4
)
=
11
2
4
x
??
+
??
??
The perimeter is given to be 13 cm. Therefore,
11
2
4
x
??
+
??
??
 =13
or
11
4
x +
 =
13
2
(dividing both sides by 2)
or x =
13 11
24
-
=
26 11 15 3
3
44 4 4
-= =
The length of the rectangle is 
3
3
4
 cm.
Example 7: The present age of Sahil’s mother is three times the present age of Sahil.
After 5 years their ages will add to 66 years. Find their present ages.
Solution: Let Sahil’s present age be x years.
         It is given that this sum is 66 years.
Therefore, 4x + 10 = 66
This equation determines Sahil’s present age which is x years. To solve the equation,
We could also choose Sahil’s age
5 years later to be x and proceed.
Why don’t you try it that way?
Sahil Mother Sum
Present age x 3x
Age 5 years later x + 5 3x + 5 4x + 10