1. By
Anuja Gupta
VIII C Roll.No.04
KV, OFD Raipur, Dehradun

2. Any equation that can be written in the form Ax + B =
C, where A, B, C are real numbers, is referred to as a linear
equation in one variable.
For example:-
5x + 15 = 35
4x + 12 = 3x + 2
(x – ½) * ½ = 1/8
3x – 6x = -12
are linear equations in one variable.

3. The linear equation may be divided in two parts the RHS
or the Right Hand Side and the LHS or the Left Hand
Side.
RHS
LHS5x + 15 = 35
Both the sides of a Linear equation are always equal.
5x + 15 35

4.  To solve a linear equation our aim is to make LHS
equal to the variable than the RHS is our solution….
As simple as that !!!
 To do this remember that Addition and subtraction are
inverse operations, and so are Division and
multiplication.
 To open a bracket use the distributive property.
 Follow the BODMAS rule.
Now let us take some examples ….

5. Problem 1:
5x + 9 = 29 find x
To make LHS equal to the variable that is ‘x’ we have to eliminate 5 and 9
To eliminate 9 subtract 9 from both sides
5x + 9 – 9 = 29 – 9 => 5x = 20
To eliminate 5 divide both sides by 5
5x/5 = 20/5 => x = 4
Now when LHS = the variable ‘x’ RHS is the answer i.e. 4
To check the validity of the answer substitute x = 4 in the original equation
5 * 4 + 9 = 20 + 9 = 29
Hence x = 4 is the correct solution.

6. Problem 2:
(2 + a) * 6 = (14 + a) 2 find ‘a’
Firstly we will open the brackets
12 +6a = 28 + 2a
To remove 2a from RHS Subtract 2a from both the sides
12 + 6a -2a = 28 + 2a – 2a => 12 + 4a = 28
To eliminate 12 Subtract 12 from both the sides
12 - 12 +4a = 28 – 12 => 4a = 16
To eliminate 4 divide both sides by 4
4a/4 = 16/4 => a = 4
Check the answer by substituting a = 4 in the original equation.

7. Problem 3:
I have some 2 rupee coins and 5 rupee coins in my pocket. The number of 2 rupee
coins is double than that of 5 rupee coins. If I have total 45 Rs. In my pocket find
the number of each type of coins.
Let I have ‘x’ coins of Rs. 5 then the number of Rs. 2 coins will be double that ‘x’
i.e. 2x. Also I have Rs. 45 in total…
=> 5 * x + 2 * 2x = 45
=> 5x + 4x = 45
=> 9x = 45
=> x = 45/9
or x = 5
That I have 5 coins of Rs. 5 and 10 coins of Rs. 2
Now check the answer
5 * 5 +2 * 10 = 25 + 20 = 45 that was what I had in my pocket.

8. Problem 4:
The length of a swimming pool is 9/4 times its width. If its perimeter
is 26 meters calculate its area.
x
Solution:
Let the length of the pool be ‘x’ meter.
Then its width will be x*9/4 or 9x/4 meter.
Perimeter of the pool is 26 meter, and the
Relation between perimeter and area is 2(l+b)
So now we have a linear equation => 2(x+9x/4) = 26
=> 2x + 9x/2 = 26
=> 4x + 9x = 52
=> 13x = 52
=> 13x/13 = 52/13
=> x = 4
Hence width of the pool is 4 meter while length is 9 meter.

9. Problem 5:
The age of father is 6 times than the son, after 12 years father will be thrice as old as
his son. Calculate their present ages.
Solution:
Present age Age after 12 years
Son x x + 12
Father 6x 6x + 12
According to the question
(x + 12) *3 = 6x + 12
=> 3x + 36 = 6x + 12
=> 3x = 24
=> x = 24/3
or x = 8
So the present age of son is 8 years and that of father is 48 years.

10. Solve:
5x + 6x = 9
7x – 2x = -25
y/12 = 5
1/3 z = 19
6/7 t = 9/5
5x – 6 = 17
-4x + 2 = 9

11.  To solve x + 3 = 7, you subtract 3 from both sides.
 To solve x - 3 = 7, what do you do?
 Addition and subtraction are inverse operations.
 We always use inverse operations to solve
equations.
Back

12. A
Project
Made by
Anuja Gupta
VIII C
Roll No. 04
KV, OFD Raipur Dehradun
I am thankful to
Wikipedia
Microsoft
Art
Anuja Gupta
Narrator
Ananya Gupta
Direction
A.K. Gupta
The End