Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
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.
The general form of linear equations in two variables x and y is
ax+by+c=0, where a=/=0, b=/=0and a,b,c are real numbers.
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 solution2x+4y-12=0 2 4 -12 lines
2x+3y-9=0 2 3 -9 = = Coincident Infinitely4x+6y-18=0 4 6 -18 lines many solutions
a1
a2
b1
b2
a1
b1
b1
c1
a1
c1
c2a2
a2 b2
b2
c2
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
are Parallel lines, then = =/=
Coincident Lines, then = =
b1 c1
a1
c1
b1
b2
c2b2
b2
a2
c2
a1
Lines of any given equation may be of three types -
Intersecting Lines Parallel Lines
Coincident Lines
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.
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
XX’
Y
Y’
(2,5)(-1,6)
(3,0)(0,1)
Unique solution X= 1 ,Y=3
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.
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
XX’
Y
Y’
(0,2)
(4,0)
(6,0)
(0,3)
No solution
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.
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
XX’
Y
Y’
(0,3)
(4.5,0)
Infinitely many solutions
(3,1)
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.
There are three algebraic methods for solving a pair of equations :-
Substitution method
Elimination method Cross-multiplication method
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.
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)
Substituting the value of x in eq.(2), we get2x-3y=12
2(-2y-1)-3y=12-4y-2-3y=12
-7y=14Y=-2
Putting the value of y in eq.(3), we getx=-2y-1
x=-2(-2)-1x=4-1x=3
Hence, the solution of the equation is (3,-2).
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)
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
2x+3y=42(5)+3y=410+3y=43y=4-10
3y=-6y= -2
Hence, the solution of the equation is (5,-2)
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
Or :- a1x + b1y = c1
a2x + b2y = c2
Then, x = y = -1
b1 c2 - b2 c1 c1a2 - c2 a1 a1b2 - a2b1
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=463x+5y=74
i.e. 2x+3y-46=0 ………(1) 3x+5y-74=0………(2)
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
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)
Related Documents
