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SOLUTION OF A LINEAR EQUATION IN TWO VARIABLES
The solution of a linear equation in two variables x and y, means a pair of values,
one for x and one for y which satisfy the given equation.
For example, in equation 3x + 2y = 12, putting the value of x = 2 and y = 3, we get
3x + 2y = (3 × 2) + (2 × 3) = 12
Hence x = 2 and y = 3 is a solution of the above equation. This solution is written as an ordered pair (2, 3).
For example, in equation 3x + 2y = 12, putting the value of x = 2 and y = 3, we get
3x + 2y = (3 × 2) + (2 × 3) = 12
Hence x = 2 and y = 3 is a solution of the above equation. This solution is written as an ordered pair (2, 3).
A linear equation in two variables has infinitely many solutions. To get many solutions
of a linear equation in two variables, a variety of values of x is chosen and put
in the linear equation in two variables to reduce it in a linear equation in one
variable, by solving which the corresponding value of y is obtained. Choosing many
different values of x and putting them in the equation many corresponding values
of y can be obtained.
Example 1:
Write four solutions for the equation 2x + y = 9.
Solution:
2x + y = 9 => y = 9 – 2x
Putting x = 0, we get y = 9 – 2(0) = 9 – 0 = 9
Putting x = 1, we get y = 9 – 2(1) = 9 – 2 = 7
Putting x = 2, we get y = 9 – 2(2) = 9 – 4 = 5
Putting x = 3, we get y = 9 – 2(3) = 9 – 6 = 3
Hence four solutions are (0, 9), (1, 7), (2, 5) and (3, 3).
Putting x = 0, we get y = 9 – 2(0) = 9 – 0 = 9
Putting x = 1, we get y = 9 – 2(1) = 9 – 2 = 7
Putting x = 2, we get y = 9 – 2(2) = 9 – 4 = 5
Putting x = 3, we get y = 9 – 2(3) = 9 – 6 = 3
Hence four solutions are (0, 9), (1, 7), (2, 5) and (3, 3).
Example 2:
Check whether (4, 0) and (1, 1) are solutions of equation x – 2y = 4 or not.
Solution:
(4, 0)
Putting x = 4 and y = 0 in equation x – 2y = 4, we get
x – 2y = 4 – 2(0) = 4
Hence (4, 0) is a solution of x – 2y = 4.
(1, 1)
Putting x = 1 and y = 1 in equation x – 2y = 4, we get
x – 2y = 1 – 2(1) = -1 which is not 4.
Hence (1, 1) is not a solution of x – 2y = 4.
Putting x = 4 and y = 0 in equation x – 2y = 4, we get
x – 2y = 4 – 2(0) = 4
Hence (4, 0) is a solution of x – 2y = 4.
(1, 1)
Putting x = 1 and y = 1 in equation x – 2y = 4, we get
x – 2y = 1 – 2(1) = -1 which is not 4.
Hence (1, 1) is not a solution of x – 2y = 4.
Example 3:
Find the value of a so that the equation 5x + 2ay = 3a may have
x =1, y = 1 as a solution.
Solution:
If x = 1, y = 1 is a solution, then
| 5(1) + 2a(1) = 3a | => | 5 + 2a = 3a |
| => | 5 = 3a – 2a | |
| => | a = 5 (Answer) |
Example 4:
Show that x = 1, y = 1 as well as x = 2, y = 5 is a solution of 4x – y – 3 = 0.
Solution:
4x – y – 3 = 0 => 4x – y = 3
(1)
Putting x = 1, y = 1 in equation (1), we get
4x – y = 4(1) – 1
= 4 – 1 = 3 (which satisfies equation(1) )
Hence x = 1, y = 1 is a solution of equation (1)
Now putting x = 2, y = 5 in equation (1), we get
4x – y = 4(2) – 5
= 8 – 5 = 3
Hence x = 2, y = 5 is a solution of equation (1).
Hence it is proved that x = 1, y = 1 as well as x = 2, y = 5 is a solution of 4x – y – 3 = 0.
Putting x = 1, y = 1 in equation (1), we get
4x – y = 4(1) – 1
= 4 – 1 = 3 (which satisfies equation(1) )
Hence x = 1, y = 1 is a solution of equation (1)
Now putting x = 2, y = 5 in equation (1), we get
4x – y = 4(2) – 5
= 8 – 5 = 3
Hence x = 2, y = 5 is a solution of equation (1).
Hence it is proved that x = 1, y = 1 as well as x = 2, y = 5 is a solution of 4x – y – 3 = 0.
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