The document discusses systems of linear equations and methods for solving them. It defines a system of linear equations as a pair of linear equations in two variables. It presents two examples of systems of linear equations. It then describes algebraic methods for solving systems, including elimination by substitution, elimination by equating coefficients, and cross multiplication. It provides examples of using elimination by substitution and elimination by equating coefficients to solve systems of two equations with two unknowns.

1. Presented By:- Abhinav Somani

2. System Of Equations
• A pair of Linear equation
in two variables is said to
form a system of linear
equation.
• For Example:-
 2x-3y+9=0
 x+4y+2=0
• These Equations form a
system of Two Linear
Equations in variables x
and y.

3. General form of linear equations in
two variables
• The general form of a linear
equation in two variables x and y
is ax + by + c = 0, a and b is not
equal to zero, where a, b and c
being real numbers.
• A solution of such an equation is
a pair of values, One for x and
the other for y, Which makes
two sides of the equation equal.
Every linear equation in two
variables has infinitely many
solutions which can be
represented on a certain line.

4. Graphical Solutions of a Linear
Equations
• Let us consider the
following system of two
simultaneous linear
equations in two variable.
 2x+y-8=0
 2x-5y+4=0
• Here we assign any value
to one of the two
variables and then
determine the value of
the other variable from
the given equation

5. Solving Equations
1. 2x+y-8=0
X 4 0
Y 0 8
2. 2x-5y+4=0
X -2 0
Y 0 4/5
• 2x+y=8
2x+0=8
x=8/4
x=2
• 2x+y=8
2x0+y=8
0+y=8
y=8
• 2x-5y= -4
2x-5x0= -4
2x-0= -4
x= -2
• 2x-5y= -4
2x0-5y= -4
0-5y= -4
y=4/5

7. Algebraic Methods of Solving
Simultaneous Linear Equations
• The most commonly used algebraic methods
of solving simultaneous linear equations in
two variables are:-
 Method of Elimination by substitution
 Method of Elimination by equating the
coefficient.
Method of Cross Multiplication.

8. Elimination by substitution
• Obtain the two equations. Let the equations be
a1x+b1y+c1=0 (i)
a2x+b2y+c2=0 (ii)
• Choose either of the two equations say (i) and
find the value of one variable say ‘y’ in terms of
x.
• Substitute the value of y, obtained in the
previous step in equation (ii) to get an equation
in x.

9. • Solve the equation obtained in the previous
step to get the value of x.
• Substitute the value of x and get the value of y.
Let us take an example
x+2y=-1 (i)
2x-3y=12 (ii)
 For (i) equation
1. x+2y=-1
x=-2y-1 (iii)

10. • Substituting the value of x in equation (ii),
we get
2x-3y=12
2(-2y-1) -3y=12
-4y-2-3y=12
-4y-3y=12+2
-7y=14
y=-2

11. • Putting the value of y in equation (iii) we get
x=-2(-2)-1
x= 4-1
x=3

12. Elimination Method
• In this method we eliminate one of the two
variables to obtain an equation in one variable
which can easily be solved.
• Putting the value of the variable in any of the
given equations. The value of the other
variable can be obtained.
• For example:-
3x+2y=11
2x+3y=4

13. • Let 3x+2y=11 (i)
2x+3y=4 (ii)
• Multiply 3 in equation (i) and 2 in equation (ii)
and subtracting equation (iv) from (iii), we get
9x+6y=33 (iii)
4x+6y=8 (iv)
5x =25
x=5

14. • Putting the value of x in equation (ii) we get,
2x+3y=4
2x5+3y=4
10+3y=4
3y=4-10
3y=-6
y=-2