Important Questions Class 9 Maths Chapter 4-Linear Equation in Two Variables

Important Questions of Class 9 Maths Chapter 4-Linear equations in two variables with solutions are available for students who are preparing 9th final exam 2020. These problems are solved by our experts, as per NCERT book formulated by CBSE board. It covers all the questions according to the syllabus which are important as per exam point of view.

Students can reach us at BYJU’S to get important questions for all chapter class 9 Maths. Practicing these extra questions will help them to revise all the concepts and get good marks in the examination.

Also Check:

• Important 2 Marks Questions for CBSE 9th Maths
• Important 3 Marks Questions for CBSE 9th Maths
• Important 4 Marks Questions for CBSE 9th Maths

Important Questions & Solutions For Class 9 Maths Chapter 4 (Linear Equation in Two Variables)

Q.1: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:

(i) x – y/5 – 10 = 0

(ii) -2x+3y = 6

(iii) y – 2 = 0

Solution:

(i) The equation x-y/5-10 = 0 can be written as:

1x + (-1/5) y + (-10) = 0

Now compare the above equation with ax + by + c = 0

Thus, we get;

a = 1

b = -⅕

c = -10

(ii) –2x + 3y = 6

Re-arranging the given equation, we get,

–2x + 3y – 6 = 0

The equation –2x + 3y – 6 = 0 can be written as,

(–2)x + 3y +(– 6) = 0

Now comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0

We get, a = –2

b = 3

c = -6

(iii) y – 2 = 0

Solution:

y – 2 = 0

The equation y – 2 = 0 can be written as,

0x + 1y + (–2) = 0

Now comparing 0x + 1y + (–2) = 0with ax + by + c = 0

We get, a = 0

b = 1

c = –2

Q.2. Write four solutions for each of the following equations:

(i) 2x + y = 7

Solution:

To find the four solutions of 2x + y = 7 we substitute different values for x and y

Let x = 0

Then,

2x + y = 7

(2×0)+y = 7

y = 7

(0,7)

Let x = 1

Then,

2x + y = 7

(2×1)+y = 7

2+y = 7

y = 7 – 2

y = 5

(1,5)

Let y = 1

Then,

2x + y = 7

(2x)+ 1 = 7

2x = 7 – 1

2x = 6

x = 3

(3,1)

Let x = 2

Then,

2x + y = 7

(2)+y = 7

4+y = 7

y = 7 – 4

y = 3

(2,3)

The solutions are (0, 7), (1,5), (3,1), (2,3)

(ii) πx + y = 9

Solution:

To find the four solutions of πx + y = 9 we substitute different values for x and y

Let x = 0

Then,

πx + y = 9

(π × 0)+y = 9

y = 9

(0,9)

Let x = 1

Then,

πx + y = 9

(π×1)+y = 9

π+y = 9

y = 9-π

(1,9)

Let y = 0

Then,

πx + y = 9

πx +0 = 9

πx = 9

x =9/π

(9/π,0)

Let x = -1

Then,

πx + y = 9

(π(-1))+y = 9

-π + y = 9

y = 9+π

(-1,9+π)

The solutions are (0,9), (1,9-π),(9/π,0),(-1,9+π)

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

Solution:

The given equation is

2x + 3y = k

According to the question, x = 2 and y = 1.

Now, Substituting the values of x and y in the equation 2x + 3y = k,

We get,

⇒(2 x 2)+ (3 × 1) = k

⇒4+3 = k

⇒7 = k

⇒k = 7

The value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k, is 7.

Q.4: Draw the graph of each of the following linear equations in two variables:

(i)y = 3x

Solution:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

y=3x

Substituting the values for x,

When x = 0,

y = 3x

y = 3

⇒ y = 0

When x = 1,

y = 3x

y = 3(1)

⇒ y = 3

The points to be plotted are (0, 0) and (1, 3)

(ii) 3 = 2x + y

Solution:

To draw a graph of linear equations in two variables, let us find out the points to plot.

To find out the points, we have to find the values which x and y can have, satisfying the equation.

Here,

3 = 2x + y

Substituting the values for x,

When x = 0,

3 = 2x + y

⇒ 3 = 2 + y

⇒ 3 = 0 + y

⇒ y = 3

When x = 1,

3 = 2x + y

⇒ 3 = 2 + y

⇒ 3 = 2 + y

⇒ y = 3 – 2

⇒ y = 1

The points to be plotted are (0, 3) and (1, 1)

Q.5: If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Solution:

The given equation is

3y = ax + 7

According to the question, x = 3 and y = 4

Now, Substituting the values of x and y in the equation 3y = ax + 7,

We get,

(3×4) = (ax3) + 7

⇒ 12 = 3a+7

⇒ 3a = 12–7

⇒ 3a = 5

⇒ a = 5/3

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The value of a, if the point (3, 4) lies on the graph of the equation 3y = ax + 7 is 5/3.

Q.6: Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.

Solution:

We have the equation,

y = 9x – 7

For A (1, 2),

Substituting the values of (x,y) = (1, 2),

We get,

2 = 9(1) – 7 = 9 – 7 = 2

For B (–1, –16),

Substituting the values of (x,y) = (–1, –16),

We get,

–16 = 9(–1) – 7 = – 9 – 7 = – 16

For C (0, –7),

Substituting the values of (x,y) = (0, –7),

We get,

– 7 = 9(0) – 7 = 0 – 7 = – 7

Hence, we find that the points A (1, 2), B (–1, –16) and C (0, –7) satisfies the line y = 9x – 7.

Hence, A (1, 2), B (–1, –16) and C (0, –7) are solutions of the linear equation y = 9x – 7

Therefore, points A (1, 2), B (–1, –16), C (0, –7) lie on the graph of linear equation y = 9x – 7.

Q.7: Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis?

Solution: According to the question,

We get the equation,

3x + 4y = 6.

We need at least 2 points on the graph to draw the graph of this equation,

Thus, the points the graph cuts

(i) x-axis

Since, the point is on x axis, we have, y = 0.

Substituting y = 0 in the equation, 3x + 4y = 6,

We get,

3x + 4×0 = 6

⇒ 3x = 6

⇒ x = 2

Hence, the point at which the graph cuts x-axis = (2, 0).

(ii) y-axis

Since, the point is on y axis, we have, x = 0.

Substituting x = 0 in the equation, 3x + 4y = 6,

We get,

3×0 + 4y = 6

⇒ 4y = 6

⇒ y = 6/4

⇒ y = 3/2

⇒ y = 1.5

Hence, the point at which the graph cuts x-axis = (0, 1.5).

Plotting the points (0, 1.5) and (2, 0) on the graph.

​Extra Questions For Class 9 Maths Chapter 4

1. If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is:
   • 2 units
   • 0 unit
2. The taxi fare in a city is as follows: For the first kilometre, the fare is ₹ 8 and for the subsequent distance it is ₹5 per km. Taking the distance covered as x km and total fare as ₹ y, write a linear equation for this information, and draw its graph.
3. Give the geometric representations of 2x+9 = 0 as an equation:
   • in one variable
   • in two variables
4. Write the linear equation such that each point on its graph has an ordinate 3 times its abscissa.