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# NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.3

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# NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables Exercise 4.3

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NCERT Solutions for class 9 Maths Chapter 4 Linear Equation in Two Variables  Exercise 4.3

by Unbeaten Ingenious
(30.6k points)

1: 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
Solution: (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 line AB as shown below.

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 have 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 below: Thus, the line PQ is required graph of x – y = 2.

(iii) y = 3x

If x = 0, then y = 3(0)
⇒ y = 0 x = 1, then y = 3(1)
⇒ y = 3 x = –1, then y = 3(–1)
⇒ y = –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 the straight line LM.
Thus, LM is the required graph of y = 3x.

Note: The graph of the equation of the form y = kx is a straight line which always passes through the origin.

( iv) 3 = 2x + y
⇒ y = 3 – 2x
∴ If x = 0, then y = 3 – 2(0)
⇒ y = 3 If x = 1, then y = 3 – 2(1)
⇒ y = 1 If x = 2, then y = 3 – (2)
⇒ y = –1

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

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

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
(2, 14) means x = 2 and y = 14
Following equations can have (2, 14) as the solution, i.e. they can pass through the point (2, 14).
(i) x + y = 16

(ii) 7x – y = 0
There can be an unlimited number of lines which can pass through the point (2, 14) because an unlimited number of lines can pass through a point.
3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
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
L.H.S. = 3y
= 3 x 4
= 12
R.H.S. = ax + 7
= a x 3 + 7
= 3a + 7
∵ L.H.S. = R.H.S.
∴ 12 = 3a + 7
or
3a = 12 – 7 = 5

or
a = (5/3)
Thus, the required value of a is (5/3)
4. 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.
Solution: Here, total distance covered = x km
Total taxi fare = ₹  y
Fare for the 1st km = ₹  8
Remaining distance = (x – 1) km
∴ Fare for (x – 1) km = ₹ 5 x (x – 1) km
Total taxi fare = ₹ 8 + ₹  5(x – 1)
∴ According to the condition,
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, y = 5(0) + 3
⇒ y = 3
When x = –1, y = 5(–1) + 3
⇒ y = –2
When x = –2, 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.
Thus, PQ is the required graph of the linear equation y = 5x + 3.

by Unbeaten Ingenious
(30.6k points)

1: 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
Solution: (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 line AB as shown below.

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 have 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 below: Thus, the line PQ is required graph of x – y = 2.

(iii) y = 3x

If x = 0, then y = 3(0)
⇒ y = 0 x = 1, then y = 3(1)
⇒ y = 3 x = –1, then y = 3(–1)
⇒ y = –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 the straight line LM.
Thus, LM is the required graph of y = 3x.

Note: The graph of the equation of the form y = kx is a straight line which always passes through the origin.

( iv) 3 = 2x + y
⇒ y = 3 – 2x
∴ If x = 0, then y = 3 – 2(0)
⇒ y = 3 If x = 1, then y = 3 – 2(1)
⇒ y = 1 If x = 2, then y = 3 – (2)
⇒ y = –1

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

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

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
(2, 14) means x = 2 and y = 14
Following equations can have (2, 14) as the solution, i.e. they can pass through the point (2, 14).
(i) x + y = 16

(ii) 7x – y = 0
There can be an unlimited number of lines which can pass through the point (2, 14) because an unlimited number of lines can pass through a point.
3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
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
L.H.S. = 3y
= 3 x 4
= 12
R.H.S. = ax + 7
= a x 3 + 7
= 3a + 7
∵ L.H.S. = R.H.S.
∴ 12 = 3a + 7
or
3a = 12 – 7 = 5

or
a = (5/3)
Thus, the required value of a is (5/3)
4. 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.
Solution: Here, total distance covered = x km
Total taxi fare = ₹  y
Fare for the 1st km = ₹  8
Remaining distance = (x – 1) km
∴ Fare for (x – 1) km = ₹ 5 x (x – 1) km
Total taxi fare = ₹ 8 + ₹  5(x – 1)
∴ According to the condition,
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, y = 5(0) + 3
⇒ y = 3
When x = –1, y = 5(–1) + 3
⇒ y = –2
When x = –2, 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.
Thus, PQ is the required graph of the linear equation y = 5x + 3.

by Unbeaten Ingenious
(30.6k points)

1: 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
Solution: (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 line AB as shown below.

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 have 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 below: Thus, the line PQ is required graph of x – y = 2.

(iii) y = 3x

If x = 0, then y = 3(0)
⇒ y = 0 x = 1, then y = 3(1)
⇒ y = 3 x = –1, then y = 3(–1)
⇒ y = –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 the straight line LM.
Thus, LM is the required graph of y = 3x.

Note: The graph of the equation of the form y = kx is a straight line which always passes through the origin.

( iv) 3 = 2x + y
⇒ y = 3 – 2x
∴ If x = 0, then y = 3 – 2(0)
⇒ y = 3 If x = 1, then y = 3 – 2(1)
⇒ y = 1 If x = 2, then y = 3 – (2)
⇒ y = –1

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

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

2. Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?
(2, 14) means x = 2 and y = 14
Following equations can have (2, 14) as the solution, i.e. they can pass through the point (2, 14).
(i) x + y = 16

(ii) 7x – y = 0
There can be an unlimited number of lines which can pass through the point (2, 14) because an unlimited number of lines can pass through a point.
3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
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
L.H.S. = 3y
= 3 x 4
= 12
R.H.S. = ax + 7
= a x 3 + 7
= 3a + 7
∵ L.H.S. = R.H.S.
∴ 12 = 3a + 7
or
3a = 12 – 7 = 5

or
a = (5/3)
Thus, the required value of a is (5/3)
4. 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.
Solution: Here, total distance covered = x km
Total taxi fare = ₹  y
Fare for the 1st km = ₹  8
Remaining distance = (x – 1) km
∴ Fare for (x – 1) km = ₹ 5 x (x – 1) km
Total taxi fare = ₹ 8 + ₹  5(x – 1)
∴ According to the condition,
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, y = 5(0) + 3
⇒ y = 3
When x = –1, y = 5(–1) + 3
⇒ y = –2
When x = –2, 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.
Thus, PQ is the required graph of the linear equation y = 5x + 3.