NCERT Class 9 Mathematics Textbook for Blind Students made Screen Readable by Professor T K Bansal.

We have seen that every linear equation in one variable has a unique solution. What can We say about the solution of 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 We substitute x = 3 and y = 2 in the above equation, We 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).

We have seen at least two solutions for 2x + 3y = 12, that is, (3, 2) and (0, 4). Can We 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. Let us Pick a value of our 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, We get

y = 8/3.

So (2, 8/3) is another solution of 2x + 3y = 12. Similarly, choosing x = − 5, We 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 the number of different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions.

Example 3

Find 4 different solutions of the equation x + 2y = 6.

Solution :

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.

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, out of the infinitely many solutions of the given equation are:
(2, 2), (0, 3), (6, 0) and (4, 1).

Remark :
Note that 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.

Example 4

Find two solutions for each of the following equations:
(i) 4x + 3y = 12
(ii) 2x + 5y = 0
(iii) 3y + 4 = 0

Solution :

(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 We take y = 0, We again get (0, 0) as a solution, which is the same as the earlier one. To get a different solution, take x = 1, say. Then We 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 0x + 3y + 4 = 0, We 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)

EXERCISE 4.2

Q1. Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions

A1. (iii), because for every value of x, there is a corresponding value of y and vice-versa.

Q2. Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y

A2.
(i) (0, 7), (1, 5), (2, 3), (4, −1)
(ii) (1, 9 − π), (0, 9), (−1, 9 + π), (9/π, 0)
(iii) (0, 0), (4, 1), (−4, 1), (2, ½)

Q3. Check which of the following are solutions of the equation x − 2y = 4 and which are not:
(i) (0, 2)
(ii) (2, 0)
(iii) (4, 0)
(iv) ( √2 , 4 √2)
(v) (1, 1)

A3.
(i) No
(ii) No
(iii) Yes
(iv) No
(v) No

Q4. Find the value of k, if x = 2, and y = 1 is a solution of the equation 2x + 3y = k.

A4. 7