2. WHAT IS LINEAR EQUATION IN ONE
VARIABLE?
A linear equation in one variable is an equation which
can be written in the form…like- (ax + b = c)
This topic deals with equations with linear expressions in
one variable..
3. let's BRIefly revise it….
An algebraic equation is an equality involving variables. It has an
equality sign. The expression on the left of the equality sign is (Left
hand side RHS). The expression on the right of the equality sign is
(Right hand side RHS)…
LIKE- 2x – 3 = 7
In this equation
2x is the variable,3 is the equality and 7 is the equation…
2x-3=LHS
7=RHS
4. MORE BRIEF CONCEPTS…
In an equation the values of the expression on the LHS and RHS are
equal. This happens to be true only for certain values of the variable.
These variable are the solutions of the equation… like-
X=5 is the solution of the question
2x=3=7. For x=5,
LHS= 2 x 5 – 3 = 7 =RHS
On the other hand x=10,is not a solution of the equation .
For x=10, LHS = 2 x 10 – 3 = 17.
So, this is not equal to RHS…
5. LINEAR EQUATIONS HAVING THE VARIABLE ONLY ON
ONE SIDE..
We can solve linear equation using the transposition method. In this
process we move a term from one side to the other side of the
equation by changing its sign.
For eg- Solve 3x – 2 = 7
Solution- 3x – 2 = 7
(ON TRANSPOSING) = 3x = 7 + 2
= 3x = 9
(ON TRANSPOSING) x = 9/3
x = 3 ( we get the result)…
RULES OF TRANSPOSITION:
+a = -a
-a = +a
xa = divide a
Divide a = xa
6. EQUATIONS HAVING THE VARIABLES ON BOTH SIDES..
We can also solve linear equations have variable on both the sides by
using the method of transposition . Lets took some examples to
understand the process for solving the equation….
Example- Solve : 4a – 3 = 2a + 7
SOLUTION- 4a -3 = 2a + 7
=4a – 2a = 7 + 3
=2a = 10
= a =10 / 2
= a = 5 ( We got the result )..
NOTE
Always gather the
variables on one side
of the equation and
the constant on the
other side of the
equation…
7. Reducing equations to simpler form….
When linear equation are in fractions then we can reduce them to simpler form by –
Taking the LCM of the denominator.
Multiply the LCM on both the sides , so that the number will reduce without the denominator
and we can solve them by the above methods…
Example- Solve the linear equation.. x/2 – 1/5 = x/3 + 1/4 + 1
SOLUTION –As the equation is in complex form, we have to reduce it ito a simper form..
1)Take LCM of the denominators 2,3,4,5 which is 60.
=30x – 12 = 20x + 15 + 60
2) Now bring all the variables on the LHS and all the constant on the RHS..
= 30x – 20 x = 15 + 12 + 60
=10x = 87
3) Dividing both sides by 10
x = 8.7 (We got the result )..
