This presentation include various methods of solving linear equations like substitution, elimination and cross-multiplication method.

3. A pair of linear equations in two variables is said to
form a system of linear equations.
For Example, 2x-3y+4=0
x+7y+1=0
Form a system of two linear equations in variables x
and y.

4. The general form of linear equations in two variables
x and y is
ax+by+c=0, where a=/=0, b=/=0
and a,b,c are real numbers.

5. Pair of lines
Comparison
of ratios
Graphical
Represen-
tation
Algebraic
Interpre-
tation
a1
a2
b1
b2 c2
c1
x-2y=0 1 -2 0 =/= Intersect- Unique
3x+4y-20=0 3 4 -20 lines solution
x+2y-4=0 1 2 -4 = =/= Parallel No solution
2x+4y-12=0 2 4 -12 lines
2x+3y-9=0 2 3 -9 = = Coincident Infinitely
4x+6y-18=0 4 6 -18 lines many
solutions
a1
a2
b1
b2
a1 b1
b1
c1
a1 c1
c2a2
a2 b2
b2
c2

6. From the table, we can observe that if the lines
represented by the equation -
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
are Intersecting lines, then =
b1
a2 b2
a1

7. are Parallel lines, then = =/=
Coincident Lines, then = =
b1 c1
a1 c1b1
b2
c2b2
b2
a2
c2
a1

9. Lines of any given equation may be of three types -
Intersecting Lines
Parallel Lines
Coincident Lines

11. Let us consider the following system of linear
equations in two variable
2x-y=-1 and 3x+2y=9
Here we assign any value to one of the two
variables and then determine the value of the
other variable from the given equation.

12. For the equation
2x-y=-1 ………(1)
2x+1=y
y =2x+1
3x+2y=9 ………(2)
2y=9-3x
9-3x
y= -------
2
x 0 2
y 1 5
x 3 -1
y 0 6

13. XX’
Y
Y’
(2,5)
(-1,6)
(3,0)(0,1)
Unique
solution
X= 1 ,Y=3

15. Let us consider the following system of linear
equations in two variable
x+2y=4 and 2x+4y=12
Here we assign any value to one of the two
variables and then determine the value of the
other variable from the given equation.

16. For the equation
x+2y=4 ………(1)
2y=4-x
y = 4-x
2
2x+4y=12 ………(2)
2x=12-4y
x = 12-4y
2
x 0 4
y 2 0
x 0 6
y 3 0

17. XX’
Y
Y’
(0,2)
(4,0)
(6,0)
(0,3)
No solution

19. Let us consider the following system of linear
equations in two variable
2x+3y=9 and 4x+6y=18
Here we assign any value to one of the two
variables and then determine the value of the
other variable from the given equation.

20. For the equation
2x+3y=9 ………(1)
3y=9-2x
y = 9-2x
3
4x+6y=18 ………(2)
6y=18-4x
y = 18-4x
6
x 0 4.5
y 3 0
x 0 3
y 3 1

21. XX’
Y
Y’
(0,3)
(4.5,0)
Infinitely
many
solutions
(3,1)

22. A pair of linear equation in two variables, which
has a unique solution, is called a consistent pair of
linear equation.
A pair of linear equation in two variables, which
has no solution, is called a inconsistent pair of
linear equation.
A pair of linear equation in two variables, which
has infinitely many solutions, is called a consistent
or dependent pair of linear equation.

24. There are three algebraic methods for solving a
pair of equations :-
Substitution method
Elimination method
Cross-multiplication method

26. Let the equations be :-
a1x + b1y + c1 = 0 ………. (1)
a2x + b2y + c2 = 0 ……….. (2)
Choose either of the two equations say (1) and find
the value of one variable, say y in terms of x.
Now, substitute the value of y obtained in the
previous step in equation (2) to get an equation in
x.

27. Solve the equations obtained in the previous step to
get the value of x. Then, substitute the value of x
and get the value of y.
Let us take an example :-
x+2y=-1 ………(1)
2x-3y=12………(2)
By eq. (1)
x+2y=-1
x= -2y-1……(3)

28. Substituting the value of x in eq.(2), we get
2x-3y=12
2(-2y-1)-3y=12
-4y-2-3y=12
-7y=14
Y=-2
Putting the value of y in eq.(3), we get
x=-2y-1
x=-2(-2)-1
x=4-1
x=3
Hence, the solution of the equation is (3,-2).

30. In this method, we eliminate one of the two
variables to obtain an equation in one variable which
can be easily solved. Putting the value of this
variable in any of the given equations, the value of
the other variable can be obtained.
Let us take an example :-
3x+2y=11……….(1)
2x+3y=4………(2)

31. Multiply 3 in eq.(1) and 2 in eq.(2) and by subtracting
eq.(4) from (3), we get
9x+6y=33…………(3)
4x+6y=08………(4)
5x=25
⇒x=5
Putting the value of x in eq.(2), we get

32. 2x+3y=4
2(5)+3y=4
10+3y=4
3y=4-10
3y=-6
y= -2
Hence, the solution of the equation is (5,-2)

34. Let the equations be :-
a1x + b1y + c1 = 0
a2x + b2y + c2 = 0
Then, x = y = 1
b1 c2 - b2 c1 c1a2 - c2 a1 a1b2 - a2b1

35. Or :-
a1x + b1y = c1
a2x + b2y = c2
Then, x = y = -1
b1 c2 - b2 c1 c1a2 - c2 a1 a1b2 - a2b1

36. In this method, we have put the values of
a1,a2,b1,b2,c1 and c2 and by solving it, we will get
the value of x and y.
Let us take an example :-
2x+3y=46
3x+5y=74
i.e. 2x+3y-46=0 ………(1)
3x+5y-74=0………(2)

37. Then, x = y = 1
x = y = 1
(3)(-74)–(5)(-46) (-46)(3)-(-74)(2) (2)(5)-(3)(3)
x = y = 1
(-222+230) (-138+148) (10-9)
b1 c2 - b2 c1 c1a2 - c2 a1 a1b2 - a2b1

38. so, x = y = 1
8 10 1
x = 1 and y = 1
8 1 10 1
x=8 and y=10
So, Solution of the equation is (8,10)