## MP Board Class 9th Maths Solutions Chapter 4 Linear Equations in Two Variables Ex 4.2

Question 1.

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.

Solution:

The true option is (iii) y = 3x + 5 has infinitely many solutions. Reason. For every value of x, there is a corresponding value of y and vice-versa.

Question 2.

Write four solutions for each of the following equations:

- 2x + y = 7
- πx + y = 9
- x = 4y

Solution:

1. 2x + y = 7

Take x = 0

2 x 0 + y = 7

∴ y = 7

First solution is (0, 7)

Take x = 1,

2 x 1 + y = 7

2 + y = 7

y = 7 – 2

∴ y = 5

Second solution is (1, 5)

Take x = 2,

2 x 2 + y = 7

2 + y = 7

y = 7 – 4

∴ y = 3

Third solution is (2, 3)

Take x = 3,

2 x 3 + y = 7

6 + y = 7

y = 1 – 6

∴ y = 1

Fourth solution is (3, 1)

2. πx + y = 9

Take y = 0,

πx + 0 = 9

πx = 9

∴ x = \(\frac{9}{π}\)

First solution is (\(\frac{9}{π}\), 0)

Take y = 1,

πx + 1 = 9

πx =9 – 1

∴ x = \(\frac{8}{π}\)

Second solution is (\(\frac{8}{π}\), 1)

Take y = 2,

πx + 2 = 9

πx = 9 – 2

∴ x = \(\frac{7}{π}\)

Third solution is (\(\frac{7}{π}\), 2)

Take y = 3,

πx + 3 = 9

πx = 9 – 3

∴ x = \(\frac{6}{π}\)

Fourth solution is (\(\frac{6}{π}\), 3)

3. x = 4y

Take x = 0,

0 = 4y

\(\frac{0}{4}\) = y

∴ y = o

First solution is (0, 0)

Take x = 4,

4 = 4y

y = \(\frac{4}{4}\) = 1

Second solution is (4, 1)

Take x = 8,

8 = 4y

y = \(\frac{8}{4}\) = 2

Third solution is (8, 2)

Take x = 12,

12 = 4y

y = \(\frac{12}{4}\) = 3

Fourth solution is (12, 3)

Question 3.

Check which of the following are solutions of the equation x – 2y = 4 and which are not:

- (0, 2)
- (2, 0)
- (4, 0)
- (\(\sqrt{2}\), 4\(\sqrt{2}\) )
- (1, 1)

Solution:

x – 2y = 4

1. Putting x = 0 and y = 2, we get

0 – 2 x 2 = 4

4 = 4

∴ (0, 2) is a solution

2. Putting x = 2 and y = 0, we get

2 – 2 x 0 = 4

2 ≠ 4

∴ (2, 0) is not the solution.

3. Putting x = 4 and y = 0, we get

4 – 2 x 0 = 4

4 = 4

∴ (4, 0) is a solution.

4. Putting x = \(\sqrt{2}\) and y = 4\(\sqrt{2}\), we get

\(\sqrt{2}\) – 2 x 4\(\sqrt{2}\) = 4

\(\sqrt{2}\) – 8\(\sqrt{2}\) = 4

– 7\(\sqrt{2}\) ≠ 4

∴ (\(\sqrt{2}\), 4\(\sqrt{2}\)) is not the solution.

5. Putting x = 1 and y = 1, we get

1 – 2 x 1 = 4

1 – 2 = 4

– 1 ≠ 4

∴ (1, 1) is not the solution.

Question 4.

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

Solution:

Putting x =2 and y – 1 in 2x + 3y – k, we get

2 x 2 + 3 x 1 = k

4 + 3 = k

∴ k = 1

Graph Of A Linear Equation In Two Variables:

The graph of a linear equation in two variables is a line. To draw a line we need atleast two points. Points are the solutions of the given equation. So to find the graph of a linear equation we will first find out three solutions and then plot these points on a suitable scale to a graph to get a line.

Steps for plotting the graph of Linear Equation in Two variables:

- Write the linear equation.
- Express y in terms of x.
- Choose three value of x and calculate the corresponding values of y from the given equation.
- Tabulate the values of x and y.
- Plot the value of x on x – axis and value of y on y – axis to a suitable scale, on a graph paper to get three points.
- Join the three points by a straight line and extend it in both the directions.
- The line obtained is the graph of the given equation.

Note:

To draw a graph of a linear equation in two variables, atleast, two solutions are required. In this chapter three solutions are taken to draw the graph for better result. Students can draw the graph by taking two solutions also.