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Made by – Group 4 
Group name- G. Cantor 
Group leader- Pragyan 
Members- Aditya 
Prashant 
Bhisham 
Anshul 
Shubham
Linear Equations 
A linear equation in two variable x and y 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 are not equal to 0. 
Example of a linear equation in two variables is 2x+4y=60
Solution of an Equation in Two Variables 
Example: 
Given the equation 2x + 3y = 18, determine if the 
ordered pair (3, 4) is a solution to the equation. 
Solution: 
We substitute 3 in for x and 4 in for y. 
2(3) + 3 (4) ? 18 
6 + 12 ? 18 
18 = 18 True. 
Therefore, the ordered pair (3, 4) is a solution to the 
equation 2x + 3y = 18.
The Rectangular Coordinate 
System 
In the rectangular coordinate system, the horizontal number line 
is the x-axis. 
The vertical number line is the y-axis. 
The point of intersection of these axes is their zero points, called 
the origin. The axes divide the plane into 4 quarters, called 
quadrants.
CARTESIAN PLANE 
Quadrant II 
( - ,+) 
Quadrant I 
(+,+) 
Quadrant IV 
(+, - ) 
Quadrant III 
( - , - ) 
y-axis 
x- axis
Plotting Points on a Graph Paper 
EXAMPLE Plot the points (3,2) and (-2,-4). 
SOLUTION
Solution for a linear equation in two variables 
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.
Graphical 
Method Example 
Solve the following system of linear equations: 
Since we are seeking out the point of intersection, we may graph the equations: 
We see here that the lines intersect each other at the point x = 2, y = 8. This is 
our solution and we may refer to it as a graphic solution to the task.
Substitution 
method 
The method of solving "by substitution" works by 
solving one of the equations (you choose which one) 
for one of the variables (you choose which one), and 
then plugging this back into the other equation, 
"substituting" for the chosen variable and solving for 
the other. Then you back-solve for the first variable.
For example if the equations are 
푥 − 2푦 = 10…… (i) and 2푥 + 푦 = 30………. (ii) 
From (i) we get 푥 = 10 + 2푦 
So now we put the value of x in equation (ii)…. 
2 10 + 2푦 + 푦 = 30 
Or, 20 + 5푦 = 30 Or, 푦 = 
10 
5 
Or, 푦 = 2 
Putting the value of y in equation (i) 
i.e. 푥 − 2 2 = 10 
or, 푥 = 10 + 4 표푟, 푥 = 14
Elimination 
In the eliMmineattiohn moedthod we eliminate 
either of the variables to solve the 
equation. 
We first make the coefficients of the 
variable equal in both the equations by 
multiplying the whole equation by an 
integer except zero. Then we eliminate the 
variable and solve the equation for the 
value of another variable.
For example if the equations are 
푥 − 2푦 = 10…… (i) and 2푥 + 푦 = 30………. (ii) 
First we make the coefficient of x in equation (i) equal to that of 
equation (ii). So, we multiply the whole equation by 2 we get, 
2푥 − 4푦 = 20 … … … (푖푖푖) 
Now we can either subtract (ii) by (iii) or vice versa. 
So, subtracting (iii) by (ii) we get, 
2푥 + 푦 − 2푥 − 4푦 = 30 − 20 
푂푟, 2푥 + 푦 − 2푥 + 4푦 = 10 
푂푟, 5푦 = 10 푖. 푒. 푦 = 
10 
5 
푖. 푒. 푦 = 2 
Now we put the value of y in equation (iii) we get, 
2푥 − 4 2 = 20 
푖. 푒. 2푥 = 20 + 8 
푖. 푒. 2푥 = 28 표푟, 푥 = 
28 
2 
표푟, 푥 = 14
Cross 
multiplication 
method 
Let’s consider the general form of a pair of linear equations a1x + b1y + c1 = 0 , and a2x + 
b2y + c2 = 0. 
When a1 divided by a2 is not equal to b1 divided by b2, the pair of linear equations will 
have a unique solution. 
To solve this pair of equations for x and y using cross-multiplication, we’ll arrange the 
variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2 as 
shown
For example if the equations are 
푥 − 2푦 = 10…… (i) and 2푥 + 푦 = 30………. (ii) 
First we write it in this form as 
푥 − 2푦 − 10 = 0 푎푛푑 2푥 + 푦 − 30 = 0 
According to the theory 
푥 
−2 − 10 
1 − 30 
= 
푦 
−10 1 
−30 2 
= 
1 
1 − 2 
2 1 
Cross multiplying we get, 
푥 
−10 − 60 
= 
푦 
−30 − −20 
= 
1 
−4 − 1 
푖푒. 
푥 
−70 
= 
푦 
−10 
= 
1 
−5 
푥 
−70 
= 
1 
−5 
푖푒. −5푥 = −70 푖푒. 푥 = − 
70 
−5 
표푟, 푥 = 14 
푦 
−10 
= 
1 
−5 
푖푒. −5푦 = −10 푖푒. 푦 = − 
10 
−5 
푖푒. 푦 = 2
Equations reducible to a 
pair of linear equations in 
two variables 
Q. Solve the pair of equations 
2 
푥 
+ 
3 
푦 
= 13 
5 
푥 
− 
4 
푦 
= −2 
Sol. Let us write the equations as 
2 
1 
푥 
+ 3 
1 
푦 
= 13 
5 
1 
푥 
− 4 
1 
푦 
= −2
We let 
1 
푥 
= 푝 푎푛푑 
1 
푦 
= 푞 
Thus the equation becomes, 
2푝 + 3푞 = 13 푎푛푑 5푝 − 4푞 = −2 
Now you can use any method to get p=2 and q=3 
Substituting the values of p and q to get, 
1 
푥 
= 2 푖푒. , 푥 = 
1 
2 
푎푛푑 
1 
푦 
= 3 푖푒. , 푦 = 
1 
3
Linear equations in two variables- By- Pragyan

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Linear equations in two variables- By- Pragyan

  • 1. Made by – Group 4 Group name- G. Cantor Group leader- Pragyan Members- Aditya Prashant Bhisham Anshul Shubham
  • 2. Linear Equations A linear equation in two variable x and y 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 are not equal to 0. Example of a linear equation in two variables is 2x+4y=60
  • 3. Solution of an Equation in Two Variables Example: Given the equation 2x + 3y = 18, determine if the ordered pair (3, 4) is a solution to the equation. Solution: We substitute 3 in for x and 4 in for y. 2(3) + 3 (4) ? 18 6 + 12 ? 18 18 = 18 True. Therefore, the ordered pair (3, 4) is a solution to the equation 2x + 3y = 18.
  • 4. The Rectangular Coordinate System In the rectangular coordinate system, the horizontal number line is the x-axis. The vertical number line is the y-axis. The point of intersection of these axes is their zero points, called the origin. The axes divide the plane into 4 quarters, called quadrants.
  • 5. CARTESIAN PLANE Quadrant II ( - ,+) Quadrant I (+,+) Quadrant IV (+, - ) Quadrant III ( - , - ) y-axis x- axis
  • 6. Plotting Points on a Graph Paper EXAMPLE Plot the points (3,2) and (-2,-4). SOLUTION
  • 7. Solution for a linear equation in two variables 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.
  • 8.
  • 9. Graphical Method Example Solve the following system of linear equations: Since we are seeking out the point of intersection, we may graph the equations: We see here that the lines intersect each other at the point x = 2, y = 8. This is our solution and we may refer to it as a graphic solution to the task.
  • 10. Substitution method The method of solving "by substitution" works by solving one of the equations (you choose which one) for one of the variables (you choose which one), and then plugging this back into the other equation, "substituting" for the chosen variable and solving for the other. Then you back-solve for the first variable.
  • 11. For example if the equations are 푥 − 2푦 = 10…… (i) and 2푥 + 푦 = 30………. (ii) From (i) we get 푥 = 10 + 2푦 So now we put the value of x in equation (ii)…. 2 10 + 2푦 + 푦 = 30 Or, 20 + 5푦 = 30 Or, 푦 = 10 5 Or, 푦 = 2 Putting the value of y in equation (i) i.e. 푥 − 2 2 = 10 or, 푥 = 10 + 4 표푟, 푥 = 14
  • 12. Elimination In the eliMmineattiohn moedthod we eliminate either of the variables to solve the equation. We first make the coefficients of the variable equal in both the equations by multiplying the whole equation by an integer except zero. Then we eliminate the variable and solve the equation for the value of another variable.
  • 13. For example if the equations are 푥 − 2푦 = 10…… (i) and 2푥 + 푦 = 30………. (ii) First we make the coefficient of x in equation (i) equal to that of equation (ii). So, we multiply the whole equation by 2 we get, 2푥 − 4푦 = 20 … … … (푖푖푖) Now we can either subtract (ii) by (iii) or vice versa. So, subtracting (iii) by (ii) we get, 2푥 + 푦 − 2푥 − 4푦 = 30 − 20 푂푟, 2푥 + 푦 − 2푥 + 4푦 = 10 푂푟, 5푦 = 10 푖. 푒. 푦 = 10 5 푖. 푒. 푦 = 2 Now we put the value of y in equation (iii) we get, 2푥 − 4 2 = 20 푖. 푒. 2푥 = 20 + 8 푖. 푒. 2푥 = 28 표푟, 푥 = 28 2 표푟, 푥 = 14
  • 14. Cross multiplication method Let’s consider the general form of a pair of linear equations a1x + b1y + c1 = 0 , and a2x + b2y + c2 = 0. When a1 divided by a2 is not equal to b1 divided by b2, the pair of linear equations will have a unique solution. To solve this pair of equations for x and y using cross-multiplication, we’ll arrange the variables x and y and their coefficients a1, a2, b1 and b2, and the constants c1 and c2 as shown
  • 15. For example if the equations are 푥 − 2푦 = 10…… (i) and 2푥 + 푦 = 30………. (ii) First we write it in this form as 푥 − 2푦 − 10 = 0 푎푛푑 2푥 + 푦 − 30 = 0 According to the theory 푥 −2 − 10 1 − 30 = 푦 −10 1 −30 2 = 1 1 − 2 2 1 Cross multiplying we get, 푥 −10 − 60 = 푦 −30 − −20 = 1 −4 − 1 푖푒. 푥 −70 = 푦 −10 = 1 −5 푥 −70 = 1 −5 푖푒. −5푥 = −70 푖푒. 푥 = − 70 −5 표푟, 푥 = 14 푦 −10 = 1 −5 푖푒. −5푦 = −10 푖푒. 푦 = − 10 −5 푖푒. 푦 = 2
  • 16. Equations reducible to a pair of linear equations in two variables Q. Solve the pair of equations 2 푥 + 3 푦 = 13 5 푥 − 4 푦 = −2 Sol. Let us write the equations as 2 1 푥 + 3 1 푦 = 13 5 1 푥 − 4 1 푦 = −2
  • 17. We let 1 푥 = 푝 푎푛푑 1 푦 = 푞 Thus the equation becomes, 2푝 + 3푞 = 13 푎푛푑 5푝 − 4푞 = −2 Now you can use any method to get p=2 and q=3 Substituting the values of p and q to get, 1 푥 = 2 푖푒. , 푥 = 1 2 푎푛푑 1 푦 = 3 푖푒. , 푦 = 1 3