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LINEAR EQUATIONS IN
TWO VARIABLES
MADE BY MANAS VIVEK MORE
CLASS 10TH B
ROLL NO. 28
INTRODUCTION
1. AN EQUATION OF THE FORM
ax + by + c = 0
2. Where a, b and c are real numbers, and a and b are both zero, is
called pair of linear equation in two variables x and y .
The standard form of pair of
linear equation in two variables x and y.
A1x + b1y + c1 = 0
 A2x + b2y +c2 = 0
 where a1 , b1, c1, a2, b2, c2, are real numbers.
LINEAR EQUATIONS
A linear equations 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 are and a and b is not equal
to 0.
An example of a linear equation in x is 2x – 3y +4 = 0 .
Definition of a linear equation
A pair of linear equations in two variables can
be solved by the :
1. Graphically method
2. Algebraic method
SUBSTITUTION METHOD
SUBSTITUTION METHOD OF SOLVING
EQUATIONS
 The graphical method of solving a equations is not always convenient specially when the
point representing the solution has non- integral coordinates. So, we have to search
alternative method of finding the solution. The algebraic method is used in this case .
 There are three types of method
1. Substitution method
2. Elimination method
3. cross multiplication method
SUBSTITUTION METHOD
STEP 1: Express y in terms of x from either equations .
STEP 2 : Substitute this value of y in the other equation resulting it to reduce in an equation resulting it to
reduce in an equation in x. solve it for x.
STEP 3 : Substitute this value of x in the equation used in the step 1 to obtain the value of y and get the value
of y.
Example
1) X + y = 17
2x -3y = 11
x + y = 17 (1)
2 x – 3y = 11 (2)
From equation (1)
y = 7 – x (3)
Substitute this value of y in equation (2) we get
2x- 3 (7 – x) = 11
2x – 21 + 3x = 11
5x – 21 = 11
5x = 11+ 21
5x = 32
x = 32/5
Substitute this value of y in equations (3)
we get y = 7- 32/5
y = 35 – 32 / 5
y = 3/5
x = 32/5
y = 3 / 5
ELEMINATION METHOD
STEP 1 : Multiply both the equation by the suitable non- zero constant to make the coefficient of one
variable numerically equal.
STEP 2 : Then add or subtract one equation from the other so that one variable gets eliminated.
STEP 3 : Solve the resulting equation in one variable so obtained and get the value .
STEP 4 : Substitute this value of x or y in either of the original equation and get the value of the other
variables .
EXAMPLE :
1) 5x +3y =70 ------------ (1)
2) 3x – 7y = 60 ------------(2)
Multiplying equations (1) by 3
Multiplying equations (2) by 5
15x + 9y =210
15x – 35y =300
Subtracting equations (4) from equation (3) we get
15x +9y =210
15x – 35y =300 44y = - 90
44y = - 90
y = - 90/44
y = - 44 / 22
Substituting this value of y in equations (1) we get
5x + 3 (- 45/22) = 70
5x – 135 / 22 = 70
5x = 70 + 135/ 22
5x = 1540 + 135/22
5x = 167/22
5x = 335/22
x = 335/22
Y = -45 / 22
CROSS – MULTIPICATION METHOD
STEP 1 : DRAW A DIAGRAM AS FOLLOWS
X Y 1
b1 c1 a1 b1
b2 c2 a2 b2
STEP 2 : THEN WRITE THE EQUATION AS FOLLOW
X Y 1
B1c2-b2c1 c1a2-C2A1 a1b2-a2b1
STEP 3 : If a1b2 – a2b1 = 0 find x and y as follows
X = ------------ y = ------------
b1c2-b2c1 c1a2-c2a1
CROSS – MULTIPICATION METHOD
1) 11x + 15y + 23 = 0
7x – 2y -20 = 0
x y 1
15 23 11 15
-2 -20 7 - 2
x = y
----------------------- -----------------------
(15) (-20) - (-2) (23) (23) (7) – (120) (11)
1
-----------------------------
(11) (-2) – (-7) (15)
CROSS – MULTIPICATION METHOD
 X y 1
----------- = --------------- ----- -----------
- 300 + 46 161 + 220 -22-105
 X = y = 1
---------- ------------ -----------
-254 381 -127
x = - 254 /- 127 = 2
y = 381 / - 127 = 3
x = 2
y = 3
GRAPHICAL METHOD OF SOLVING
EQUATION
 Graphical representation of linear equation in two variables is a straight line. As a straight line consist of an
infinite no of points lying on it .
 Given a pair of linear equation in two variables representing two straight lines taken together , only one of
the following three possibilities can occur .
 The two lines intersect at one point .
 The two lines are parallel .
 The two lines are coincident.
The two lines interests at one point . The two are parallel The two lines are Coincident
EXAMPLE
SANGEETA HAS SOCKS AND KERCHIEFS WHICH ARE TOGETHER 40 IN NO. IF SHE HAS 5 LESS
HANDKERCHIEFS AND 5 MORE SOCKS , THE NUMBER OF SOCKS BECOMES FOUR TIMES THE NUMBER OF
HAND KERCHIEFS. REPRESENT THIS SITUATION IN ALGERBRAICALLY AND GRAPHICALLY .
X + Y = 40
X + 5 = 4(Y – 5)
X + 5 = 4Y – 20
X – 4Y = -25
FOR EQUATION (1)
X + Y = 40
Y = 40 – X
FOR EQUATION (2)
X- 4Y = -25
4Y = X + 25
X 20 40
Y 20 0
X 15 55
Y 10 20
Linear equations in two variables
Linear equations in two variables

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Linear equations in two variables

  • 1. LINEAR EQUATIONS IN TWO VARIABLES MADE BY MANAS VIVEK MORE CLASS 10TH B ROLL NO. 28
  • 2. INTRODUCTION 1. AN EQUATION OF THE FORM ax + by + c = 0 2. Where a, b and c are real numbers, and a and b are both zero, is called pair of linear equation in two variables x and y . The standard form of pair of linear equation in two variables x and y. A1x + b1y + c1 = 0  A2x + b2y +c2 = 0  where a1 , b1, c1, a2, b2, c2, are real numbers.
  • 3. LINEAR EQUATIONS A linear equations 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 are and a and b is not equal to 0. An example of a linear equation in x is 2x – 3y +4 = 0 . Definition of a linear equation A pair of linear equations in two variables can be solved by the : 1. Graphically method 2. Algebraic method
  • 4. SUBSTITUTION METHOD SUBSTITUTION METHOD OF SOLVING EQUATIONS  The graphical method of solving a equations is not always convenient specially when the point representing the solution has non- integral coordinates. So, we have to search alternative method of finding the solution. The algebraic method is used in this case .  There are three types of method 1. Substitution method 2. Elimination method 3. cross multiplication method
  • 5. SUBSTITUTION METHOD STEP 1: Express y in terms of x from either equations . STEP 2 : Substitute this value of y in the other equation resulting it to reduce in an equation resulting it to reduce in an equation in x. solve it for x. STEP 3 : Substitute this value of x in the equation used in the step 1 to obtain the value of y and get the value of y. Example 1) X + y = 17 2x -3y = 11 x + y = 17 (1) 2 x – 3y = 11 (2) From equation (1) y = 7 – x (3) Substitute this value of y in equation (2) we get 2x- 3 (7 – x) = 11 2x – 21 + 3x = 11 5x – 21 = 11 5x = 11+ 21 5x = 32 x = 32/5 Substitute this value of y in equations (3) we get y = 7- 32/5 y = 35 – 32 / 5 y = 3/5 x = 32/5 y = 3 / 5
  • 6. ELEMINATION METHOD STEP 1 : Multiply both the equation by the suitable non- zero constant to make the coefficient of one variable numerically equal. STEP 2 : Then add or subtract one equation from the other so that one variable gets eliminated. STEP 3 : Solve the resulting equation in one variable so obtained and get the value . STEP 4 : Substitute this value of x or y in either of the original equation and get the value of the other variables . EXAMPLE : 1) 5x +3y =70 ------------ (1) 2) 3x – 7y = 60 ------------(2) Multiplying equations (1) by 3 Multiplying equations (2) by 5 15x + 9y =210 15x – 35y =300 Subtracting equations (4) from equation (3) we get 15x +9y =210 15x – 35y =300 44y = - 90 44y = - 90 y = - 90/44 y = - 44 / 22 Substituting this value of y in equations (1) we get 5x + 3 (- 45/22) = 70 5x – 135 / 22 = 70 5x = 70 + 135/ 22 5x = 1540 + 135/22 5x = 167/22 5x = 335/22 x = 335/22 Y = -45 / 22
  • 7. CROSS – MULTIPICATION METHOD STEP 1 : DRAW A DIAGRAM AS FOLLOWS X Y 1 b1 c1 a1 b1 b2 c2 a2 b2 STEP 2 : THEN WRITE THE EQUATION AS FOLLOW X Y 1 B1c2-b2c1 c1a2-C2A1 a1b2-a2b1 STEP 3 : If a1b2 – a2b1 = 0 find x and y as follows X = ------------ y = ------------ b1c2-b2c1 c1a2-c2a1
  • 8. CROSS – MULTIPICATION METHOD 1) 11x + 15y + 23 = 0 7x – 2y -20 = 0 x y 1 15 23 11 15 -2 -20 7 - 2 x = y ----------------------- ----------------------- (15) (-20) - (-2) (23) (23) (7) – (120) (11) 1 ----------------------------- (11) (-2) – (-7) (15)
  • 9. CROSS – MULTIPICATION METHOD  X y 1 ----------- = --------------- ----- ----------- - 300 + 46 161 + 220 -22-105  X = y = 1 ---------- ------------ ----------- -254 381 -127 x = - 254 /- 127 = 2 y = 381 / - 127 = 3 x = 2 y = 3
  • 10. GRAPHICAL METHOD OF SOLVING EQUATION  Graphical representation of linear equation in two variables is a straight line. As a straight line consist of an infinite no of points lying on it .  Given a pair of linear equation in two variables representing two straight lines taken together , only one of the following three possibilities can occur .  The two lines intersect at one point .  The two lines are parallel .  The two lines are coincident. The two lines interests at one point . The two are parallel The two lines are Coincident
  • 11. EXAMPLE SANGEETA HAS SOCKS AND KERCHIEFS WHICH ARE TOGETHER 40 IN NO. IF SHE HAS 5 LESS HANDKERCHIEFS AND 5 MORE SOCKS , THE NUMBER OF SOCKS BECOMES FOUR TIMES THE NUMBER OF HAND KERCHIEFS. REPRESENT THIS SITUATION IN ALGERBRAICALLY AND GRAPHICALLY . X + Y = 40 X + 5 = 4(Y – 5) X + 5 = 4Y – 20 X – 4Y = -25 FOR EQUATION (1) X + Y = 40 Y = 40 – X FOR EQUATION (2) X- 4Y = -25 4Y = X + 25 X 20 40 Y 20 0 X 15 55 Y 10 20