1. B.J.P.S Samiti’s
M.V.HERWADKAR ENGLISH MEDIUM HIGH SCHOOL
CLASS 8th LINEAR EQUATION IN ONE VARIABLE
Program:
Semester:
Course: NAME OF THE COURSE
Staff Name: VINAYAK PATIL 1
2. WHAT IS ALGEBRAIC EXPRESSION
• Algebraic expression is the expression having constants
and 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.
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 2
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)7y+9=16
• (ii) 2.5 = z / 0.6
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 3
4. WHAT IS LINEAR EQUATION IN ONE
VARIABLE
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 4
•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.
•Example: 3x+2= -4
5. Solving Equations which have Linear Expressions
on one Side and Numbers on the other Side
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 5
• 4x−7=9
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.
7. Word Problem on Simple Linear equation in one
variable
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 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
8. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 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
9. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 9
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.
10. Solving Equations having the Variable on
both Sides
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 10
Example: 5m−12=6−m
Solution: Transposing 12 to RHS and m to
LHS
5m+m=6+12
or,6m=18
Dividing both the sides by 6
m=3
11. Reducing Equations to Simpler Form
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 11
• 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.
12. Solve the linear equation
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 12
• 8x + 4 = 3(x – 1) + 7
Solution: Given,8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7
8x + 4 = 3x + 4
8x – 3x = 4 – 4
5x = 0
x = 0/5
x = 0
13. M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 13
• 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
14. • 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 a = 7
Therefore, another number 5a = 5 x 7 = 35 Thus, required numbers
are 7 and 35
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 14
15. HOMEWORK
• 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.
M.V.HERWADKAR ENGLISH MEDIUM SCHOOL 15
