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2. IMPLICIT EDU

3. MATHEMATICS LINEAR
EQUATIONS IN 1
VARIABLE

4. TODAY, WE ARE GOING TO KNOW
THESE POINTS… KEEP AN EYE ON
THE LEARNING OUTCOMES:-
LINEAR EQUATIONS IN ONE VARIABLE
LAWS OF EQUALITY
METHODS OF SOLVING LINEAR EQUATIONS
APPLICATIONS OF LINEAR EQUATIONS
WORD PROBLEMS ON LINEAR EQUATIONS
LET’S BEGIN …

5. LINEAR EQUATIONS IN ONE
VARIABLE
Two Algebraic Expressions connected by the sign of equality is
called an Equation.
The expression on the left of the equality sign is called the Left
Hand Side (LHS), while that on the right of the equality sign is called
Right Hand Side (RHS).
If the expressions are in 1 variable and the highest degree of that
variable is 1, the equation is called LINEAR EQUATION IN ONE
VARIABLE.
General form of LINEAR EQUATION IN ONE VARIABLE is :-
ax+b=c , where a, b and c are constants and a is a non-zero
number.
For example, 5x+10=15 , here a=5, b=10, and c=15; x is a
LHS
RHS

6. LAWS OF EQUALITY
To balance a linear equation at each step, the following rules must be
followed:-
1) Add a term to both the sides
2) For e.g. --- 4x-12=0 => (4x-12)+12=0+12 => 4x=12
3) Subtract a term from both sides
4) For e.g. --- 9y+17=9 => (9y+17)-17=9-17 => 9y=(-
8)
5) Multiply by a non-zero term on both the sides
6) For e.g. --- ⅘ x=4 => (⅘ x)*5=4*5 =>
4x=20
7) Divide by a non-zero term on both the sides

7. METHODS OF SOLVING LINEAR
INEQUATIONS
Some of the Methods of Solving a Linear Equation are:-
 ELIMINATION METHOD
 For example, 5x+3= 2(2x+4) => 5x+3=4x+8 … (using Distributive
property)
 => (5x+3)-4x=(4x+8)- 4x … (Subtracting 4x from both sides)
 => x+3=8 => (x+3)-3=8-3 … (Subtracting 3 from both sides)
 => x=5 …
This is the Process of using LAWS OF EQUALITY for finding the value of
variables.

8. METHODS OF SOLVING LINEAR
INEQUATIONS
 TRANSPOSITION METHOD
 For example, 1.2 x+ 3.6= 2.4 x + 6
=> 1.2 x+3.6-6=2.4x … (Transposing 6 from RHS to LHS)
=> 1.2x-2.4 = 2.4x … (As, 3.6-6=2.4)
=> 1.2x=2.4x + 2.4 … (Transposing 2.4 from LHS to RHS)
=> 1.2x-2.4x=2.4 … (Transposing 2.4x from RHS to LHS)
=> -1.2x=2.4 … (As, 1.2-2.4=(-1.2))
=> x= 2.4/(-1.2) … (Transposing (-1.2) from LHS to RHS)
THEREFORE, x= (-2) … (As, 2.4/(-1.2)=(-2)) …

9. LET’S HAVE A LOOK AT 1 WORD
PROBLEM :-
 0.6 of a number is 20 less than the original number. Find the number.
 Answer :- Let the unknown number be n.
 According to the Question, 0.6 n= n-20
 => 0.6 n-n= (-20) => -0.4 n= (-20)
 Therefore, n= (-20)/(-0.4) = 50 ...
I have made this question using TRANSPOSITION METHOD.
You can also do this using Elimination Method.
The Result and Equation would be same.
We will get back with more interesting questions. Thank You !
Goodbye !

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