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Unit-4(Examples)

Linear Equations in Two Variables

**Example-1 :-** Write each of the following equations in the form ax + by + 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

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

(i) x = –5 can be written as 1.x + 0.y = –5, or 1.x + 0.y + 5 = 0. (ii) y = 2 can be written as 0.x + 1.y = 2, or 0.x + 1.y – 2 = 0. (iii) 2x = 3 can be written as 2x + 0.y – 3 = 0. (iv) 5y = 2 can be written as 0.x + 5y – 2 = 0.

**Example-3 :-** Find four different solutions of the equation x + 2y = 6.

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).

**Example-4 :-** Find two solutions for each of the following equations:

(i) 4x + 3y = 12

(ii) 2x + 5y = 0

(iii) 3y + 4 = 0

(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).

**Example-5 :-** Given the point (1, 2), find the equation of a line on which it lies. How many such equations are there?

Here (1, 2) is a solution of a linear equation you are looking for. So, you are looking for any line passing through the point (1, 2). One example of such a linear equation is x + y = 3. Others are y – x = 1, y = 2x, since they are also satisfied by the coordinates of the point (1, 2). In fact, there are infinitely many linear equations which are satisfied by the coordinates of the point (1, 2).

**Example-6 :-** Draw the graph of x + y = 7.

The graph of x + y = 7 So, we can use the following table to draw the graph:

**Example-7 :-** You know that the force applied on a body is directly proportional to the acceleration produced in the body. Write an equation to express this situation and plot the graph of the equation.

Here the variables involved are force and acceleration. Let the force applied be y units and the acceleration produced be x units. From ratio and proportion, you can express this fact as y = kx, where k is a constant. (From your study of science, you know that k is actually the mass of the body.) Now, since we do not know what k is, we cannot draw the precise graph of y = kx. However, if we give a certain value to k, then we can draw the graph. Let us take k = 3, i.e., we draw the line representing y = 3x. For this we find two of its solutions, say (0, 0) and (2, 6). From the graph, you can see that when the force applied is 3 units, the acceleration produced is 1 unit. Also, note that (0, 0) lies on the graph which means the acceleration produced is 0 units, when the force applied is 0 units.

**Example-8 :-** For each of the graphs given in Fig. 4.5 select the equation whose graph it is from the choices given below:

(a)

(i) x + y = 0 (ii) y = 2x (iii) y = x (iv) y = 2x + 1

(b)

(ii) x + y = 0 (ii) y = 2x (iii) y = 2x + 4 (iv) y = x – 4

(c)

(iii) x + y = 0 (ii) y = 2x (iii) y = 2x + 1 (iv) y = 2x – 4

(a) In Figure, the points on the line are (–1, –2), (0, 0), (1, 2). By inspection, y = 2x is the equation corresponding to this graph. We find that the y-coordinate in each case is double that of the x-coordinate. (b) In Figure, the points on the line are (–2, 0), (0, 4), (1, 6). We know that the coordinates of the points of the graph (line) satisfy the equation y = 2x + 4. So, y = 2x + 4 is the equation corresponding to the graph in Figure. (c) In Figure, the points on the line are (–1, –6), (0, –4), (1, –2), (2, 0). By inspection, you can see that y = 2x – 4 is the equation corresponding to the given graph (line).

**Example-9 :-** Solve the equation 2x + 1 = x – 3, and represent the solution(s) on

(i) the number line,

(ii) the Cartesian plane.

We solve 2x + 1 = x – 3, to get 2x – x = –3 – 1 i.e., x = –4 (i) The representation of the solution on the number line is shown in Figure, where x = – 4 is treated as an equation in one variable. (ii) We know that x = – 4 can be written as x + 0.y = – 4 which is a linear equation in the variables x and y. This is represented by a line. Now all the values of y are permissible because 0.y is always 0. However, x must satisfy the equation x = – 4. Hence, two solutions of the given equation are x = – 4, y = 0 and x = – 4, y = 2.

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