NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Question

Our NCERT Solutions Class 9 Maths Chapter 4 Pair of Linear Equations in Two Variables will help you understand and clear all the NCERT question-answer doubts. We have covered all the topics and provide solutions to all in-text and exercise questions. One must refer to our solution for one-stop solutions for all concepts Linear Equations in Two Variables. Our NCERT solutions are designed by an expert in the subject matter who has a very good understanding of all the concepts in this field. Experts have used different kinds of methods to make these solutions easy to understand for the users’ using methods like diagrams, different equations, graphs, Shortcuts, tips, and tricks.

• Linear Equations in Two Variables
• Converting statements into Equations and drawing graphs of linear equations.
• Types of Graphs made by Pair of Linear Equations in Two Variables- Two lines that are Intersecting, two lines that are parallel, Coincident Lines.
• Consistency of equations by finding the ratio of a1/a2, b1/b2, and c1/c2 to check whether lines Intersecting Lines (unique solution), Coincident Lines (infinitely many solutions), and Parallel Lines (no solutions).
• Solutions of Equations from graphs.
• Solving a pair of Linear Equations by Substitution Method, Elimination Method, and Cross Multiplication method.
• Solve statement questions with the help of the given word problem.
• Solve complicated equations with the help of substituting variable method.

Our NCERT Solutions Class 9 Maths Chapter 4 Pair of Linear Equations in Two Variable will help students comprehend the concepts with ease. Our NCERT solution is best for last-minute preparation for their examination. To score good marks one must completely go through all the solutions provided here.

Answer 1

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

1. 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)

Answer:

Let the cost of a notebook to be = ₹ x

Let the cost of a pen to be = ₹ y

According to the question,

The cost of a notebook is twice the cost of a pen.

i.e., Cost of a notebook = 2 × Cost of a pen

x = 2 × y

x = 2y

x - 2y = 0

x - 2y = 0 is the linear equation in two variables to represent the statement ‘The cost of a notebook is twice the cost of a pen’.

2. 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) 2x + 3y = \(9.\overline{35}\) 

(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

Answer:

(i) 2x + 3y = \(9.\overline{35}\) 

2x + 3y = \(9.\overline{35}\) 

Re-arranging the equation, we get,

2x + 3y - \(9.\overline{35}\) = 0

The equation 2x + 3y - \(9.\overline{35}\) = 0 can be written as,

2x + 3y + (- \(9.\overline{35}\)) = 0

Now comparing 2x + 3y + (- \(9.\overline{35}\)) = 0 with ax + by + c = 0

We get,

a = 2

b = 3

c = \(-9.\overline{35}\) 

(ii) x – (y/5) – 10 = 0

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

1x + (-1/5)y +( –10) = 0

Now comparing x + (-1/5)y + (–10) = 0 with ax + by + c = 0

We get,

a = 1

b = -(1/5)

c = -10

(iii) –2x + 3y = 6

–2x + 3y = 6

Re-arranging the 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

(iv) x = 3y

x = 3y

Re-arranging the equation, we get,

x - 3y = 0

The equation x - 3y = 0 can be written as,

1x + (-3)y + (0)c = 0

Now comparing 1x + (-3)y + (0)c = 0 with ax + by + c = 0

We get, a = 1

b = -3

c = 0

(v) 2x = –5y

2x = –5y

Re-arranging the equation, we get,

2x + 5y = 0

The equation 2x + 5y = 0 can be written as,

2x + 5y + 0 = 0

Now comparing 2x+5y+0= 0 with ax + by + c = 0

We get,

a = 2

b = 5

c = 0

(vi) 3x + 2 = 0

3x + 2 = 0

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

3x + 0y + 2 = 0

Now comparing 3x + 0 + 2= 0 with ax + by + c = 0

We get, 

a = 3

b = 0

c = 2

(vii) y – 2 = 0

y – 2 = 0

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

0x + 1y + (–2) = 0

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

We get,

a = 0

b = 1

c = –2

(viii) 5 = 2x

5 = 2x

Re-arranging the equation, we get,

2x = 5

i.e., 2x  –5 = 0

The equation 2x – 5 = 0 can be written as,

2x + 0y – 5 = 0

Now comparing 2x + 0y – 5 = 0 with ax + by + c = 0

We get,

a = 2

b = 0

c = -5

3. Which one of the following options is true, and why?

y = 3x + 5 has

1. A unique solution
2. Only two solutions
3. Infinitely many solutions

Answer:

Let us substitute different values for x in the linear equation y = 3x + 5,

x	0	1	2	...	100
y, where y = 3x + 5	5	8	11	...	305

From the table, it is clear that x can have infinite values, and for all the infinite values of x, there are infinite values of y as well.

Hence, (iii) infinitely many solutions is the only option true.

4. Write four solutions for each of the following equations:

(i) 2x + y = 7

(ii) πx + y = 9

(iii) x = 4y

Answer:

(i) 2x + y = 7

When x = 0, 2(0) + y = 7 ⇒ y = 7

∴ Solution is (0, 7)

When x =1, 2(1) + y = 7 ⇒ y = 7 – 2 ⇒ y = 5

∴ Solution is (1, 5)

When x = 2, 2(2) + y =7y = 7 – 4 ⇒ y = 3

∴ Solution is (2, 3)

When x = 3, 2(3) + y = 7y = 7 – 6 ⇒ y = 1

∴ Solution is (3, 1).

(ii) πx + y = 9

When x = 0, π(0) + y = 9 ⇒ y = 9 – 0 ⇒ y = 9

∴ Solution is (0, 9)

When x = 1, π(1) + y = 9 ⇒ y = 9 – π

∴ Solution is (1, (9 – π))

When x = 2, π(2) + y = 9 ⇒ y = 9 – 2π

∴ Solution is (2, (9 – 2π))

When x = -1,π(-1) + y = 9 ⇒ y = 9 + π

∴ Solution is (-1, (9 + π))

(iii) x = 4y

When x = 0, 4y = 1 ⇒ y = 0

∴ Solution is (0, 0)

When x = 1, 4y = 1 ⇒ y = 1/4

∴ Solution is (1,1/4)

When x = 4, 4y = 4 ⇒ y = 1

∴ Solution is (4, 1)

When x = 4, 4y = 4 ⇒ y = -1

∴ Solution is (-4, -1)

Answer 2

5. 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)

Answer:

(i) (0,2) means x = 0 and y = 2

Puffing x = 0 and y = 2 in x – 2y = 4, we get

L.H.S. = 0 – 2(2) = -4.

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

∴ x = 0, y = 2 is not a solution.

(ii) (2, 0) means x = 2 and y = 0

Putting x = 2 and y = 0 in x – 2y = 4, we get

L.H:S. 2 – 2(0) = 2 – 0 = 2.

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

∴ (2,0) is not a solution.

(iii) (4, 0) means x = 4 and y = 0

Putting x = 4 and y = o in x – 2y = 4, we get

L.H.S. = 4 – 2(0) = 4 – 0 = 4 =R.H.S.

∴ L.H.S. = R.H.S.

∴ (4, 0) is a solution.

(iv) (√2, 4√2) means x = √2 and y = 4√2

Putting x = √2 and y = 4√2 in x – 2y = 4, we get

L.H.S. = √2 – 2(4√2) = √2 – 8√2 = -7√2

But R.H.S. = 4

∴ L.H.S. ≠ R.H.S.

∴ (√2 , 4√2) is not a solution.

(v) (1, 1) means x =1 and y = 1

Putting x = 1 and y = 1 in x – 2y = 4, we get

LH.S. = 1 – 2(1) = 1 – 2 = -1. But R.H.S = 4

∴ LH.S. ≠ R.H.S.

∴ (1, 1) is not a solution.

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

Answer:

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 × 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.

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

(i) x + y = 4

(ii) x – y = 2

(iii) y = 3x

(iv) 3 = 2x + y

Answer:

(i) x + y = 4

⇒ y = 4 – x

If we have x = 0, then y = 4 – 0 = 4

x = 1, then y =4 – 1 = 3

x = 2, then y = 4 – 2 = 2

∴ We get the following table:

Plot the ordered pairs (0, 4), (1,3) and (2,2) on the graph paper. Joining these points, we get a straight line AB as shown.

Thus, the line AB is the required graph of x + y = 4

(ii) x – y = 2 

⇒ y = x – 2

If we have x = 0, then y = 0 – 2 = -2

x = 1, then y = 1 – 2 = -1

x = 2, then y = 2 – 2 = 0

∴ We get the following table:

Plot the ordered pairs (0, -2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of x – y = 2

(iii) y = 3x

If we have x = 0,

then y = 3(0) ⇒ y = 0

x = 1, then y = 3(1) = 3

x = -1, then y = 3(-1) = -3

∴ We get the following table:

Plot the ordered pairs (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM as shown.

Thus, the line LM is the required graph of y = 3x.

(iv) 3 = 2x + y

⇒ y = 3 – 2x

If we have x = 0, then y = 3 – 2(0) = 3

x = 1,then y = 3 – 2(1) = 3 – 2 = 1

x = 2,then y = 3 – 2(2) = 3 – 4 = -1

∴ We get the following table:

Plot the ordered pairs (0, 3), (1, 1) and (2, – 1) on the graph paper. Joining these points, we get a straight line CD as shown.

Thus, the line CD is the required graph of 3 = 2x + y.

8. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer:

(2, 14) means x = 2 and y = 14

Equations which have (2,14) as the solution are (i) x + y = 16, (ii) 7x – y = 0

There are infinite number of lines which passes through the point (2, 14), because infinite number of lines can be drawn through a point.

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

Answer:

The equation of the given line is 3y = ax + 7

∵ (3, 4) lies on the given line.

∴ It must satisfy the equation 3y = ax + 7

We have,

(3, 4) ⇒ x = 3 and y = 4.

Putting these values in given equation, we get

3 x 4 = a x 3 + 7

⇒ 12 = 3a + 7

⇒ 3a = 12 – 7 = 5

⇒ a = 5/3

Thus, the required value of a is 5/3.

10. The taxi fare In a city Is as follows: For the first kilometre, the fare Is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs. y, write a linear equation for this Information, and draw Its graph.

Answer:

Here, total distance covered = x km and total taxi fare = Rs. y

Fare for 1km = Rs. 8

Remaining distance = (x – 1) km

∴ Fare for (x – 1)km = Rs.5 x(x – 1)

Total taxi fare = Rs. 8 + Rs. 5(x – 1)

According to question,

y = 8 + 5(x – 1) = y = 8 + 5x – 5

⇒ y = 5x + 3,

which is the required linear equation representing the given information.

Graph: We have y = 5x + 3

When x = 0, then y = 5(0) + 3 ⇒ y = 3

x = -1, then y = 5(-1) + 3 ⇒ y = -2

x = -2, then y = 5(-2) + 3 ⇒ y = -7

∴ We get the following table:

Now, plotting the ordered pairs (0, 3), (-1, -2) and (-2, -7) on a graph paper and joining them, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation y = 5x + 3.

Answer 3

11. From the choices given below, choose the equation whose graphs are given in Fig. (1) and Fig. (2).

For Fig. (1)

(i) y = x
(ii) x + y = 0
(iii) y = 2x
(iv) 2 + 3y = 7x

For Fig. (2)

(i) y = x + 2
(ii) y = x – 2
(iii) y = -x + 2
(iv) x + 2y = 6

Answer:

For Fig. (1),

The correct linear equation is x + y = 0

[As (-1, 1) = -1 + 1 = 0 and (1, -1) = 1 + (-1) = 0]

For Fig.(2),

The correct linear equation is y = -x + 2

[As (-1, 3) 3 = -1(-1) + 2 = 3 = 3 and (0, 2)

⇒ 2 = -(0) + 2 ⇒ 2 = 2]

12. 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

(i) 2 units

(ii) 0 unit

Answer:

Constant force is 5 units.

Let the distance travelled = x units and work done = y units.

Work done = Force x Distance

⇒ y = 5 × x

⇒ y = 5x

For drawing the graph, we have y = 5x

When x = 0, then y = 5(0) = 0

x = 1, then y = 5(1) = 5

x = -1, then y = 5(-1) = -5

∴ We get the following table:

Ploffing the ordered pairs (0, 0), (1, 5) and (-1, -5) on the graph paper and joining the points, we get a straight line AB as shown.

From the graph, we get

(i) Distance travelled = 2 units i.e., x = 2

∴ If x = 2, then y = 5(2) = 10

⇒ Work done = 10 units.

(ii) Distance travelled = 0 unit i.e., x = 0

∴ If x = 0 ⇒ y = 5(0) – 0

⇒ Work done = 0 unit.

13. Yamini and Fatima, two students of Class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs.xand Rs.y.) Draw the graph of the same.

Answer:

Let the contribution of Yamini = Rs. x

and the contribution of Fatima Rs. y

∴ We have x + y = 100 ⇒ y = 100 – x

Now, when x = 0, y = 100 – 0 = 100

x = 50, y = 100 – 50 = 50

x = 100, y = 100 – 100 = 0

∴ We get the following table:

Plotting the ordered pairs (0,100), (50,50) and (100, 0) on a graph paper using proper scale and joining these points, we get a straight line PQ as shown.

Thus, the line PQ is the required graph of the linear equation x + y = 100.

14. In countries like USA and Canada, temperature is measured In Fahrenheit, whereas in countries like India, it is measured in Celsius. Here Is a linear equation that converts Fahrenheit to Celsius:

F = (\(\frac95\))C + 32

(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.

(ii) If the temperature Is 30°C, what is the temperature in Fahrenheit?

(iii) If the temperature is 95°F, what is the temperature in Celsius?

(iv) If the temperature is 0°C, what Is the temperature In Fahrenheit and If the temperature is 0°F, what Is the temperature In Celsius?

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find It.

Answer:

(i) We have

F = (\(\frac95\))C + 32

When C = 0 , F = (\(\frac95\)) x 0 + 32 = 32

When C = 15, F = (\(\frac95\))(-15) + 32= -27 + 32 = 5

When C = -10, F = \(\frac95\) (-10)+32 = -18 + 32 = 14

We have the following table:

Plotting the ordered pairs (0, 32), (-15, 5) and (-10,14) on a graph paper. Joining these points, we get a straight line AB.

(ii) From the graph, we have 86°F corresponds to 30°C.

(iii) From the graph, we have 95°F corresponds 35°C.

(iv) We have, C = 0

From (1), we get

F = (\(\frac95\))0 + 32 = 32

Also, F = 0

From (1), we get

0 = (\(\frac95\))C + 32

⇒ \(\frac{-32\times 5}9\) = C

⇒ C = -17.8

(v) When F = C (numerically)

From (1), we get

F = \(\frac95\)F + 32 ⇒ F – \(\frac95\)F = 32

⇒ \(\frac{-4}5\)F = 32 ⇒ F = -40

∴ Temperature is –40° both in F and C.

Answer 4

15. Give the geometric representations of y = 3 as an equation

(i) in one variable
(ii) in two variables

Answer:

(i) y = 3

∵ y = 3 is an equation in one variable, i.e., y only.

∴ y = 3 is a unique solution on the number line as shown below:

(ii) y = 3

We can write y = 3 in two variables as 0.x + y = 3

Now, when x = 1, y = 3

x = 2, y = 3

x = -1, y = 3

∴ We get the following table:

Plotting the ordered pairs (1, 3), (2, 3) and (-1, 3) on a graph paper and joining them, we get a line AB as solution of 0.

x + y = 3,

i.e. y = 3.

16. Give the geometric representations of 2x + 9 = 0 as an equation

(i) in one variable
(ii) in two variables

Answer:

(i) 2x + 9 = 0

We have, 2x + 9 = 0

⇒ 2x = – 9

⇒ x = \(\frac{-9}2\)

which is a linear equation in one variable i.e., x only.

Therefore, x = \(\frac{-9}2\) is a unique solution on the number line as shown below:

(ii) 2x + 9=0

We can write 2x + 9 = 0 in two variables as 2x + 0, y + 9 = 0

or \(x = \frac{-9-0.y}{2}\)

∴ When y = 1, \(x = \frac{-9-0(1)}{2} = -\frac92\) 

Thus, we get the following table:

Now, plotting the ordered pairs \((\frac{-9}2,3)\), \((\frac{-9}2,3)\) and \((\frac{-9}2,3)\) on a graph paper and joining them, we get a line PQ as solution of 2x + 9 = 0.