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

Let us first recall what we have studied so far. Let us consider the following equation:
2x + 5 = 0
Its solution, that is, the root of the equation, is −2.5. This can be represented on the number line as shown below:

Note : For the sake of convenience, throughout this chapter, the captions of the figures are given above the figure themselves and the alt text is provided with each and every figure. I hope you will enjoy this lesson, happy learning!

Figure 4.1

While solving an equation, we must always keep the following points in mind:
The solution of a linear equation is not affected when:
(i) the same number is added to (or subtracted from) both sides of the equation.
(ii) We multiply or divide both the sides of the equation by the same non-zero number.

Let us now consider the following situation:
In a One-day International Cricket match between India and Sri Lanka played in Nagpur, two Indian batsmen together scored 176 runs. Express this information in the form of an equation.
Here, we can see that the score of neither of the two players is known, that is , there are two unknown quantities. Let us use x and y to denote these variables. So, the number of runs scored by one of the batsmen is x, and the number of runs scored by the other is y. We know that
x + y = 176,
which is the required equation.

This is an example of a linear equation in two variables. It is customary to denote the variables in such equations by the letters x and y, however other letters may also be used. Some examples of linear equations in two variables are:
1.2s + 3t = 5,
p + 4q = 7,
π u + 5v = 9 and
3 = √2 x − 7y.

Note that we can put these equations in the form
1.2s + 3t − 5 = 0,
p + 4q − 7 = 0,
πu + 5v − 9 = 0 and
√2 x − 7y − 3 = 0, respectively.

So, any equation which can be put in the form a x + b y + c = 0, where a, b and c are real numbers, and a, b are not both 0, is called a linear equation in two variables. This means that we can think of many and many such equations.

Example 1

Write each of the following equations in the form of a x + b y + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 4.37
(ii) x − 4 = √3 y
(iii) 4 = 5x − 3y
(iv) 2x = y

Solution :

(i) 2x + 3y = 4.37 can be written as
2x + 3y − 4.37 = 0.
Here a = 2, b = 3 and c = − 4.37.

(ii) The equation x − 4 = √3 y can be written as
x − √ 3 y −4 = 0.
Here a = 1, b = − 3 and c = −4.

(iii) The equation 4 = 5x −3y can be written as
5x −3y −4 = 0.
Here a = 5, b = −3 and c = −4.

Do We agree that it can also be written as −5x + 3y + 4 = 0 ?
In this case a = −5, b = 3 and c = 4.

(iv) The equation 2x = y can be written as
2x − y + 0 = 0.
Here a = 2, b = −1 and c = 0.

Equations of the type ax + b = 0 are also examples of linear equations in two variables because they can be expressed as

A x + 0 y + b = 0

For example, 4 − 3x = 0 can be written as
−3x + 0 y + 4 = 0.

Example 2

Write each of the following as an equation in two variables:
(i) x = −5
(ii) y = 2
(iii) 2x = 3
(iv) 5y = 2

Solution :

(i) x = −5 can be written as
1x + 0y = −5, or
1x + 0y + 5 = 0.

(ii) y = 2 can be written as
0x + 1y = 2, or
0x + 1y −2 = 0.

(iii) 2x = 3 can be written as
2x + 0y −3 = 0.

(iv) 5y = 2 can be written as
0x + 5y −2 = 0.

EXERCISE 4.1

Note : For sake of convenience the answer to each question is given just after the question itself. The letters Q represents the question and A represents the corresponding answer.

Q1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y).

A1. x − 2y = 0

Q2. Express the following linear equations in the form a x + b y + c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35 (with Bar on 5)
(ii) x − y/5 − 10 = 0
(iii) −2x + 3y = 6
(iv) x = 3y
(v) 2x = −5y
(vi) 3x + 2 = 0
(vii) y − 2 = 0
(viii) 5 = 2x

A2.
(i) 2x + 3y − 9.35 (bar on 5) = 0; a = 2, b = 3, c = −9.35 (with dash on 5)

(ii) x − y/5 − 10 = 0; a = 1, b = −1/5, c = −10

(iii) −2x + 3y − 6 = 0; a = −2, b = 3, c = −6

(iv) 1.x − 3y + 0 = 0; a = 1, b = −3, c = 0

(v) 2x + 5y + 0 = 0; a = 2, b = 5, c = 0

(vi) 3x + 0y + 2 = 0; a = 3, b = 0, c = 2

(vii) 0x + 1y − 2 = 0; a = 0, b = 1, c = −2

(viii) −2x + 0y + 5 = 0; a = −2, b = 0, c = 5