CHAPTER 2 MATHS CLASS 8

1. LINEAR EQUATION IN
ONE VARIABLE
CLASS VIII

2.  Algebraic expression is the expression having
constants and variable. It can have multiple
variable and multiple power of the variable.
 We already know the below terms from previous
class. Lets recall those terms.
What is equation?
An equation is a condition on a variable.
What is variable?
A variable takes on different numerical values;
its value is not fixed. Variables are denoted
usually by letters of the alphabets, such as x, y, z,
l, m, n, p, etc.
What is algebraic expression

3.  An algebraic equation is an equality involving
variables. It says that the value of the
expression on one side of the equality sign is
equal to the value of the expression on the
other side.
 Example(i)7x-9=16
 (ii) 1.5 =y / 1.6

4.  We will restrict the above equation with two
conditions
a) algebraic equation in one variable
b) variable will have power 1 only
 Example(i)5x=25
(ii)2x 3=92x
−
 Linear Equation is an equation which is in the
form of ax+b=0,ax=b etc. where a and b is real
number. x is variable of power one.
What is Linear equation in one Variable

5.  2x 3=5
−
3x 11=22
−
How to solve Linear equation in one Variable
Transpose (changing the side of the number) the
numbers to the side where all number are
present. We know the sign of the number
changes when we transpose it to other side.
 Now you will have an equation have variable on
one side and number on other side. Add/subtract
on both the side to get single term.
 Now divide or multiply on both the side to get the
value of the variable.
Solving Equations which have Linear Expressions
on one Side and Numbers on the other Side

6.  Solution: Transposing 3 to other side
2x=5+3
or,2x=8
Dividing both the sides by 2
x=4
 Ex2. y + 3 = 10
Solution:
 Given y+ 3 = 10
 By subtracting 3 from both sides, we get
 y + 3 – 3 = 10 – 3
 ⇒ y = 10 – 3
 ⇒ y = 7
1.Example:2x 3=5
−

7.  Linear equation can be used to solve many word
Problem. The procedure is simple
First read the problem carefully. Write down the
unknown and known
 Assume one of the unknown to x and find the other
unknown in term of that
 Create the linear equation based on the condition given
 Solve them by using the above method
 Example: When five is added to three more than a
certain number, the result is 19. What is the
 number? Sol: Let certain number be x.According to the
given condition we have 3x+5=19 now we can solve for x.
Word Problem on Simple Linear equation in one
variable

8.  Example: The perimeter of a rectangular swimming pool is 154 m. Its
length is 2 m more than twice its breadth. What are the length and the
breadth of the pool?
Solution: The unknown are length and breadth.
Let the breadth be x m.
Then as per question the length will be (2x + 2) m.
Perimeter of swimming pool = 2(l + b) = 154 m
or,2(2x+2+x)=154
or,2(3x+2)=154
Dividing both sides by 2
3x+2=77
Transposing 2 to R.H.S, we obtain
3x=77 2
−
or,3x=75
Dividing 3 on both the sides
x=25
So, Breadth is 25 m
Length =2x + 2 = 2 × 25 + 2 = 52m
Hence, the breadth and length of the pool are 25 m and 52 m respectively.

9. Example: 2x 3=6 x
− −
Solution: Transposing 3 to RHS and x to LHS
2x+x=6+3
or,3x=9
Dividing both the sides by 3
x=3.Ans
Solving Equations having the Variable on both
Sides
:

10.  You will find many situations where the linear
equation may be having number in denominator.
We can perform the below steps to simplify them
and solve it.
 Take the LCM of the denominator of both the LHS
and RHS.
 Multiple the LCM on both the sides, this will
reduce the number without denominator and we
can solve using the method described above.
Reducing Equations to Simpler
Form

11.  8x + 4 = 3(x – 1) + 7
 Solution:
 Given,8x + 4 = 3(x – 1) + 7
By removing bracket from RHS, we get
8x + 4 = 3x – 3 + 7
By transposing 3x to LHS, we get
8x-3x + 4 = -3 + 7
⇒ 8x – 3x + 4 = 4
By transposing 4 to RHS, we get
8x – 3x = 4 – 4
⇒ 5x = 0
After dividing both sides by 5, we get
or, 5x/5=0/5
or, X=0
Solve the linear equation

12.  Ques:A positive number is 5 times another number. If 21 is added to
both the numbers, then one of the new numbers becomes twice the
other new number. What are the numbers?
 Solution: Let the given positive number = a
 Therefore, another number which is 5 times of it = 5a
 Now, after adding 21 to both of the number .First number = a + 21
 Second number = 5a + 21
 According to question, one new number becomes twice of the other
new number.Therefore,Second number = 2 x first number
 i.e. 5a + 21 = 2 (a + 21)
 ⇒ 5a + 21 = 2a + 42
 By transposing ‘2a’ to LHS, we get
 ⇒ 5a + 21 – 2a = 42
 Now, after transposing 21 to RHS, we get
 ⇒ 5a – 2a = 42 – 21
 ⇒ 3a = 21
 After dividing both sides by 3, we get
 Therefore, another number 5a = 5 x 7 = 35
 Thus, required numbers are 7 and 35

13. Solve the equation:
(5x +3)/4 - (2x -4)/3=5
Or,[(5x +3)/4] x 12 – [(2x -4)/3] x 12= 5 x12
Or,3(5x +3) - 4(2x -4) = 60
Or, 15x + 9 -8x +16 =60
Or, 7x +25 =60
Or, 7x = 60-25
Or, 7x = 35
Or, x= 35/7
Or, x= 5

14. Solve the following question:
1. Find the solution of 3x-4 = 12
2. Solve: 5x-9 = 8
3. What should be subtracted from thrice the rational number -8/3 to get
5/2?
4. The sum of three consecutive multiples of 7 is 63. Find these multiples.
5. Solve 3x/4 – 7/4 = 5x + 12
6. Perimeter of a rectangle is 13cm. if its width is 11/4 cm, find its length.
7. The present age of Seta’s father is three times the present age of Sita.
After six years sum of their ages will be 69 years. Find their present ages.
8. The digits of a two-digit number differ by 3. If digits are interchanged and
the resulting number is added to the original number, we get 121. Find the
original number.
9. (x-2)/(x+1) = ½. Find x
10. Sanjay will be 3 times as old as he was 4 years ago after 18 years. Find
his present age.
11. If the sum of two numbers is 30 and their ratio is 2/3 then find the
numbers.
12. The numerator of a fraction is 2 less than the denominator. If one is
added to its denominator, it becomes 1/2 find the fraction.

15. Thank you