Linear Equations in two variables-NCERT Solutions

Class IX Math
NCERT Solution for Linear Equations in Two Variables

NCERT QUESTIONS WITH SOLUTIONS

EXERCISE: 4.1

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 Rs. x and that of a pen to be Rs. y).

Sol: Let the cost of a notebook = Rs x

The cost of a pen = y

According to the condition, we have

[Cost of a notebook] = 2 × [Cost of a pen]

i.e. [x] = 2 × [Y]

or x = 2y

or x – 2y = 0

Thus, the required linear equation is × – 2y = 0.

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)

(ii)

(iii) –2x + 3y = 6 (iv) x = 3y

(v) 2x = –5y (vi) 3x + 2 = 0 (vii) y – 2 = 0 (viii) 5 = 2x

Sol: (i) We have

Comparing it with ax + bx + c = 0, we have a = 2, b = 3 and

(ii) We have

Comparing with ax + bx + c = 0, we get

Note: Above equation can also be compared by:

Multiplying throughout by 5,

or 5x – y – 50 = 0

or 5(x) + (–1)y + (–50) = 0

Comparing with ax + by + c = 0, we get a = 5, b = –1 and c = –50.

(iii) We have –2x + 3y = 6

⇒ –2x + 3y – 6 = 0

⇒ (–2)x + (3)y + (–6) = 0

Comparing with ax + bx + c = 0, we get a = –2, b = 3 and c = –6.

(iv) We have x = 3y

x – 3y = 0

(1)x + (–3)y + 0 = 0

Comparing with ax + bx + c = 0, we get a = 1, b = –3 and c = 0.

(v) We have 2x = –5y

⇒ 2x + 5y =0

⇒ (2)x + (5)y + 0 = 0

Comparing with ax + by + c = 0, we get a = 2, b = 5 and c = 0.

(vi) We have 3x + 2 = 0

⇒ 3x + 2 + 0y = 0

⇒ (3)x + (10)y + (2) = 0

Comparing with ax + by + c = 0, we get a = 3, b = 0 and c = 2.

(vii) We have y – 2 = 0

⇒ (0)x + (1)y + (–2) = 0

Comparing with ax + by + c = 0, we have a = 0, b = 1 and c = –2.

(viii) We have 5 = 2x

⇒ 5 – 2x = 0

⇒ –2x + 0y + 5 = 0

⇒ (–2)x + (0)y + (5) = 0

Comparing with ax + by + c = 0, we get a = –2, b = 0 and c = 5.

EXERCISE: 4.2

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

Sol: Option (iii) is true because a linear equation has an infinitely many solutions.

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

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

Sol: (i) 2x + y = 7

When x = 0, 2(0) + y = 7

⇒ 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 = 7

⇒ y = 7 – 4

⇒ y = 3

∴ Solution is (2, 3).

When x = 3, 2(3) + y = 7

⇒ y = 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 × = 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 × = –1, π(–1) + y = 9

⇒ – π + y = 9

⇒ y = 9 + π

∴ Solution is {–1, (9 + π)}

(iii) x = 4y

When x = 0, 4y = 0

⇒ y = 0

∴ Solution is (0, 0).

When x = 1, 4y = 1

⇒ y = 0

∴ Solution is (0, 0)

When x = 4, 4y = 4

⇒

3. 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)

(v) (1, 1)

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

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

L.H.S. = 0 – 2(2) = –4

But R.H.S. = 4

L.H.S. ≠ R.H.S.

∴ x = 0, y = 0 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 = 0 in x – 2y = 4, we get

L.H.S. = 4 – 2(0) = 4 – 0 = 4

But R.H.S. = 4

L.H.S. ≠ R.H.S.

∴ (4, 0) is a solution.

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

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

L.H.S. = 1 – 2(1) = 1 – 2 = –1

But R.H.S. = 4

⇒ L.H.S ≠ R.H.S.

∴ (1, 1) is not a solution.

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

Sol: We have 2x + 3y = k

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

2(2) + 3(1) = k

⇒ 4 + 3 = k

⇒ 7 = k

Thus, the required value of k = 7.

EXERCISE: 4.3

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

Sol: (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(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–ry

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

Sol: (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.

Sol: 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. R.H.S. = ax + 7

= 3 × 4 = a × 3 + 7

= 12 = 3a + 7

L.H.S. = R.H.S.

12 = 3a + 7

3a = 12 – 7 = 5

Thus, the required value of a is

4. 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 y, write a linear equation for this information and draw its graph.

Sol: Here, total distance covered = x km

Total taxi fare = y

Fare for the 1st km = Rs. 8

Remaining distance = (x – 1) km

∴ Fare for (x – 1) km = Rs. 5 × (x – 1) km

Total taxi fare = Rs. 8 + Rs. 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 × = 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.

5. From the choices given below, choose the equation whose graphs are given in Fig. (i) and Fig. (ii).

For Fig. (i) For Fig. (ii)

(i) y = x (i) y = x + 2

(ii) x + y = 0 (ii) y = × – 2

(iii) y = 2x (iii) y = x + 2

(iv) 2 + 3y = 7x (iv) x + 2y = 6

Sol: For Fig. (i), the correct linear equation is x + y = 0

[∴ (–1, 1) ⇒ –1 + 1 = 0 and (1, –1) = 1 + (–1) = 0]

For Fig. (ii), the correct linear equation is y = –x + 2

[∴ (–1, 3) ⇒ 3 = –(–1) + 2 ⇒ 3 = 3 and (0, 2) 2 = –(0) + 2 ⇒ 2 = 2]

6. 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

Sol: Constant force is 5 units. Let the distance travelled = x units and work done = y units.

Since, Work done = Force × Displacement

⇒ y = 5 × x

⇒ y = 5x

Drawing the graph

We have y = 5x

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

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

When x = 1.5, then y = 5(1.5) = 7.5

∴ We get the following table:

Plotting the ordered pairs (0, 0), (1, 5) and (1.5, 7.5) on the graph paper and joining the points, we get a straight line OB.

From the graph, we get

(i) Distance travelled = 2 units

∴ x = 2, then y = 10 units

⇒ Work done = 10 units.

(ii) Distance travelled = 0 units

∴ y = 5x y = 5(0)= 0

∴ Work done = 0 unit.

7. Yamini and Fatima. two students of Class IX of a school, together contributed 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 X and y.) Draw the graph of the same.

Sol: 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

When x = 50, y = 100 – 50 = 50

When x = 100, y = 100 – 100 = 0

We get the following table:

For drawing the graph, plot the ordered pairs (0, 100), (50, 50) and (t UU, U) on a grapn paps t. Joining these points we get a line PQ.

8. 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.