Please Wait... While Loading Full Video### LINEAR EQUATIONS IN TWO VARIABLES FOR CBSE-NCERT

You have seen that every linear equation in one variable has a unique solution.

In a linear equation involving two variables, As there are two variables in the equation, a solution means a pair of values, one for x and one for y which satisfy the given equation.

Let us consider the equation `2x + 3y = 12`. Here, `x = 3` and `y = 2` is a solution because when you substitute `x = 3` and `y = 2` in the equation above, you find that

`2x + 3y = (2 × 3) + (3 × 2) = 12`

This solution is written as an ordered pair `(3, 2)`, first writing the value for `x` and then the value for y. Similarly, (0, 4) is also a solution for the equation above.

On the other hand, `(1, 4)` is not a solution of `2x + 3y = 12`, because on putting `x = 1` and `y = 4` we get `2x + 3y = 14`, which is not `12`. Note that `(0, 4)` is a solution but not `(4, 0)`.

You have seen at least two solutions for 2x + 3y = 12, i.e., (3, 2) and (0, 4). Can you find any other solution? Do you agree that (6, 0) is another solution?

Verify the same. In fact, we can get many many solutions in the following way. Pick a value of your choice for x (say x = 2) in 2x + 3y = 12. Then the equation reduces to 4 + 3y = 12,

which is a linear equation in one variable. On solving this, you get ` y = 8/3 ` . So ` ( 2, 8/3) ` is another solution of 2x + 3y = 12. Similarly, choosing x = – 5, you find that the equation becomes `–10 + 3y = 12.`

This gives ` y = (22)/3 ` . So ` (-5, (22)/3 ) ` is another solution of `2x + 3y = 12`. So there is no end to different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions.

In a linear equation involving two variables, As there are two variables in the equation, a solution means a pair of values, one for x and one for y which satisfy the given equation.

Let us consider the equation `2x + 3y = 12`. Here, `x = 3` and `y = 2` is a solution because when you substitute `x = 3` and `y = 2` in the equation above, you find that

`2x + 3y = (2 × 3) + (3 × 2) = 12`

This solution is written as an ordered pair `(3, 2)`, first writing the value for `x` and then the value for y. Similarly, (0, 4) is also a solution for the equation above.

On the other hand, `(1, 4)` is not a solution of `2x + 3y = 12`, because on putting `x = 1` and `y = 4` we get `2x + 3y = 14`, which is not `12`. Note that `(0, 4)` is a solution but not `(4, 0)`.

You have seen at least two solutions for 2x + 3y = 12, i.e., (3, 2) and (0, 4). Can you find any other solution? Do you agree that (6, 0) is another solution?

Verify the same. In fact, we can get many many solutions in the following way. Pick a value of your choice for x (say x = 2) in 2x + 3y = 12. Then the equation reduces to 4 + 3y = 12,

which is a linear equation in one variable. On solving this, you get ` y = 8/3 ` . So ` ( 2, 8/3) ` is another solution of 2x + 3y = 12. Similarly, choosing x = – 5, you find that the equation becomes `–10 + 3y = 12.`

This gives ` y = (22)/3 ` . So ` (-5, (22)/3 ) ` is another solution of `2x + 3y = 12`. So there is no end to different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions.

Q 3270267116

Find four different solutions of the equation `x + 2y = 6`.

Class 9 Chapter 4 Example 3

Class 9 Chapter 4 Example 3

By inspection, `x = 2, y = 2` is a solution because for `x = 2, y = 2`

`x + 2y = 2 + 4 = 6`

Now, let us choose x = 0. With this value of x, the given equation reduces to 2y = 6

which has the unique solution y = 3. So x = 0, y = 3 is also a solution of x + 2y = 6.

Similarly, taking y = 0, the given equation reduces to x = 6. So, x = 6, y = 0 is a solution

of x + 2y = 6 as well. Finally, let us take y = 1. The given equation now reduces to

x + 2 = 6, whose solution is given by x = 4. Therefore, (4, 1) is also a solution of the

given equation. So four of the infinitely many solutions of the given equation are:

`(2, 2), (0, 3), (6, 0) and (4, 1)`.

An easy way of getting a solution is to take x = 0 and get the corresponding value of y. Similarly, we can put y = 0 and obtain the corresponding value of x.

Q 3280267117

Find two solutions for each of the following equations:

(i) `4x + 3y = 12`

(ii) `2x + 5y = 0`

(iii) `3y + 4 = 0`

Class 9 Chapter 4 Example 4

(i) `4x + 3y = 12`

(ii) `2x + 5y = 0`

(iii) `3y + 4 = 0`

Class 9 Chapter 4 Example 4

(i) Taking `x = 0`, we get `3y = 12`, i.e., `y = 4`. So, `(0, 4)` is a solution of the

given equation. Similarly, by taking `y = 0`, we get `x = 3`. Thus, `(3, 0) ` is also a solution.

(ii) Taking x = 0, we get 5y = 0, i.e., y = 0. So (0, 0) is a solution of the given equation.

Now, if you take y = 0, you again get (0, 0) as a solution, which is the same as the

earlier one. To get another solution, take x = 1, say. Then you can check that the

corresponding value of y is ` - 2/5` . So ` ( 1 , - 2/5 ) ` is another solution of `2x + 5y = 0`.

(iii) Writing the equation 3y + 4 = 0 as 0.x + 3y + 4 = 0, you will find that `y = - 4/3` for

any value of x. Thus, two solutions can be given as ` ( 0 , - 4/3) ` and ` ( 1 , - 4/3)` .