1.
Linear Equations
Definition of a Linear Equation
A linear equation in two variable x is an
equation that can be written in the form ax +
by + c = 0, where a ,b and c are real numbers
and a and b is not equal to 0.
An example of a linear equation in x is .

2.
Let ax + by +c = O , where a ,b , c are real numbers
such that a and b ≠ O. Then, any pair of values of x
and y which satisfies the equation ax + by +c = O, is
called a solution of it.
A Pair of Linear Equation In Two Variables
Can Be Solved –
i. By Graphical Method
ii. By Algebraic Method
SOLUTIONS OF A LINEAR EQUATION

3.
GRAPHICAL SOLUTIONS OF A LINEAR EQUATION
Let us consider the following system of linear
equations in two variable
I. 2x-y=-1
II. 3x+2y=9
Now solution of each equation are to be taken by-
i. Firstly express one variable in terms of other.
ii. Now assign any value to one variable and then
determine the value of other variable.
Then plot the equations and determine the solutions
Of system of linear equation in two variables

4.
For Eq. 1
2x - y= -1
y = 2x + 1
Put x = 0
y = 2(0) + 1 y = 1
Put x = 2
y = 2(2) + 1 y = 5
For Eq. 2
3x + 2y = 9
y = 9 - 3x/2
Put x = 3
y = 9 - 3(3) /2 y = 0
Put x = -1
y = 9 - 3(-1) /2 y = 6
X 0 2
y 1 5
X 3 -1
y 0 6

5.
1 2 3 4-4 -3 -2 -1
5
4
3
2
1
-1
-2
-3
-4
-5
-6
( 2,5 )
( -1,6 )
( 3,0 )
( 0,1 )
X = 1
Y = 3
Is the
solution
for system
of
equations

6.
TYPES OF SOLUTIONS of system of equations
UNIQUE
SOLUTION
consistent
a1/a2 ≠ b1/b2
Lines intersecting at
a point
INFINITE
SOLUTIONS
consistent
a1/a2 = b1/b2 = c1/c2
Coincident Lines
NO
SOLUTION
Non
consistent
a1/a2 = b1/b2 ≠ c1/c2 Parallel Lines

7.
TYPES OF METHOD:-
TO SOLVE A PAIR OF LINEAR
EQUATION IN TWO VARIABLE
BY ALGEABRIC METHODS
I. Elimination Method
II.Substitution Method
III.Cross-Multiplication Method

8.
• Make one variable equal in both the equations
and then either add or subtract the equations
to eliminate the variable.
• The resulting equation in one variable is
solved and then by substituting this value of
the variable in either of the given equation,
the value of the other variable is also
obtained.
ELIMINATION METHOD

9.
Q. Solve using the method of Elimination
I. 2x + y = 8
II. 2(x + 6y = 15)
For making the coefficients of x in eq.(1) and eq.(2) equal,
we multiply eq.(2) by 2 and get
2x + y = 8 -(3)
2x + 12y = 30 -(4)
Subtracting Eq.(3) from eq.(4), we have
11y = 22
y = 2
Put this value of y in eq.(1)
2x + 2 = 8
2x = 6 x = 3

10.
• Find the value of one variable in the terms of
other variable.
Substitute it in other equation and we will get
value of one of the variable.
Then put the value of variable in any one of the
equation and the value of other variable is also
obtained.
SUBSTITUTION METHOD

11.
Q. Solve using the method of Substitution
I. x + 2y = -1
II. 2x - 3y = 12
From Eq.(1), we have
x = -1 – 2y - (3)
Substituting this value of x in eq.(2), we have
2(-1 – 2y) – 3y =12
- 2 - 4y – 3y = 12
- 7y = 14 y = -2
Put this value of y in Eq.(3), we have
x = -1 -2(-2)
x = -1 + 4 x = 3

12.
CROSS- MULTIPLICATION METHOD

13.
Q. Solve using the method of cross-Multiplication
5x + 3y = 35 5x + 3y - 35 = 0
2x + 4y = 28 2x + 4y - 28 = 0
a1 = 5 b1 = 3 c1= -35
a2 = 2 b2 = 4 c2= -28
___x___ = ___y___ = ___1___
(3)(-28)-(4)(-35) (-35)(2)-(-28)(5) (5)(4)-(2)(3)
-84 + 140 -70 + 140 20 – 6
56/14 70/14 14
x = 4 and y = 5