Class 9 - Maths - Linear Equations In 2 Variables

Exercise 4.1

Question 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).

Answer:

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

=> x = 2 * y

=> x = 2y

=> x - 2y = 0

This is the required linear equation.

Question 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 = 9.35                      (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 = 9.35

=> 2x + 3y – 9.35 = 0

=> 2x + 3y + (-9.35) = 0

Compare with ax + by + c = 0, we get

a = 2, b = 3, c = -9.35

(ii) x – y/5 – 10 = 0

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

Compare with ax + by + c = 0, we get

a = 1, b = -1/5, c = -10

(iii) –2x + 3y = 6

=> -2x + 3y – 6 = 0

=> -2x + 3y + (-6) = 0

Compare with ax + by + c = 0, we get

a = -2, b = 3, c = -6

(iv) x = 3y

=> x – 3y = 0

=> x + (-3)y + 0 = 0

Compare with ax + by + c = 0, we get

a = 1, b = -3, c = 0

(v) 2x = –5y

=> 2x + 5y = 0

=> 2x + 5y + 0 = 0

Compare with ax + by + c = 0, we get

a = 2, b = 5, c = 0

(vi) 3x + 2 = 0

=> 3x + 0*y + 2 = 0

Compare with ax + by + c = 0, we get

a = 3, b = 0, c = 0

(vii) y – 2 = 0

=> 0*x + 1*y + (-2) = 0

Compare with ax + by + c = 0, we get

a = 0, b = 1, c = -1

(viii) 5 = 2x

=> -2x + 5 = 8

=> -2x + 0 * y + 5 = 0

Compare with ax + by + c = 0, we get

a = -2, b = 0, c = 5

Exercise 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

Answer:

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

linear equation in two variables has many solutions. We keep changing the value of x and solve

the linear equation for the corresponding value of y.

Question 2:

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

=> 0 + y = 7

=> y =7

So, the solution is (0, 7).

When x = 1,

2(1) + y = 7

=> y = 7 – 2

=> y = 5

So, the solution is (1, 5).

When x = 2,

2(2) + y = 7

=> y = 7 – 4

=> y = 3

So, the solution is (2, 3).

When x = 3,

2(3) + y = 7

=> y = 7 – 6

=> y = 1

So, the solution is (3, 1).

(ii) πx + y = 9

When x = 0

π(0) + y = 9

=> y = 9 – 0

=> y = 9

So, the solution is (0, 9).

When × = 1, π(1) + y = 9

=> y = 9 – π

So, the solution is {1, (9 – π)}

When x = 2,

π(2) + y = 9

=> y = 9 – 2π

So, the solution is {2, (9 – 2π)}

When × = –1,

π(–1) + y = 9

=> -π + y = 9

=> y = 9 + π

So, the solution is {–1, (9 + π)}

(iii) x = 4y

When x = 0, 4y = 0

=> y = 0

So, the solution is (0, 0).

When x = 1, 4y = 1

=> y = 0

So, the solution is (0, 0)

When x = 4,

4y = 4

=> y = 4/4

=> y = 1

So, the solution is (4, 1)

When x = -4,

4y = -4

=> y = -4/4

=> y = -1

So, the solution is (4, -1)

Question 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) (√2, 4√2)                (v) (1, 1)

Answer:

Given, equation is: x – 2y = 4

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

LHS: x – 2y = 0 – 2 * 2

= 0 – 4

= -4 ≠ RHS

So, (0, 2) is not the solution of given equation.

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

LHS: x – 2y = 2 – 2 * 0

= 2 – 0

= 2 ≠ RHS

So, (2, 0) is not the solution of given equation.

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

LHS: x – 2y = 4 – 2 * 0

= 4 – 0

= 4 = RHS

So, (4, 0) is the solution of given equation.

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

LHS: x – 2y = √2 – 2 * 4√2

= √2 – 8√2

= -6√2 ≠ RHS

So, (√2, 4√2) is not the solution of given equation.

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

LHS: x - 2y = 1 – 2 * 1

= 1 – 2

= -1 ≠ RHS

So, (1, 1) is not the solution of given equation.

Question 4:

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

Answer:

Given, x = 2, y = 1 is a solution of the equation 2x + 3y = k

=> 2 * 2 + 3 * 1 = k

=> 4 + 3 = k

=> k = 7

So, the value of k is 7

Exercise 4.3

Question 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 Answer:

(i) x + y = 4

For x = 0, y = 4

For x = 1, y = 2

So, we have points (0, 4) and (1, 3)

(ii) x – y = 2

For x = 0, y = -2

For x = 1, y = -1

So, we have points (0, -2) and (1, -1)

(iii) y = 3x

For x = 0, y = 0

For x = 1, y = 3

So, we have points (0, 0) and (1, 3)

(iv) 3 = 2x + y

For x = 0, y = 3

For x = 1, y = 1

So, we have points (0, 3) and (1, 1)

Question 2:

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

Following equations can have (2. 14) as the solution, i.e. they can pass through the point (2, 4).

(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.

Question 3:

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

Answer:

Given, the point (3, 4) lies on the graph of the equation 3y = ax + 7

=> 3 * 4 = a * 3 + 7

=> 12 = 3a + 7

=> 3a = 12 – 7

=> 3a = 5

=> a = 5/3

So, the value of a is 5/3

Question 4:

The taxi fare in a city is as follows: For the first kilometre, the fare is Rs 8 and for the sub-sequent 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:

Total fare is calculated as

y = 1 * 8 + (x - 1)5, where x is in km and y is in Rs.

So, the required equation is:

y = 8 + 5x – 5

=> y = 5x + 3

Now, for x = 1

y = 5 * 1 + 3

=> y = 5 + 3

=> y = 8

For x = 2

y = 5 * 2 + 3

=> y = 10 + 3

=> y = 13

So, the points are (1, 8) and (2, 13). Question 5:

From the choices given below, choose the equation whose graphs are given in Fig. 4.6 and Fig. 4.7.

For Fig. 4. 6                                                                 For Fig. 4.7

(i) y = x                    (ii) x + y = 0                              (i) y = x + 2                   (ii) y = x – 2

(iii) y = 2x              (iv) 2 + 3y = 7x                         (iii) y = –x + 2               (iv) x + 2y = 6 Answer:

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

[Since (–1, 1) => -1 + 1 = 0 and (1, –1) = 1 + (–1) = 0]

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

[Since (-1, 3) => 3 = -(-1) + 2 => 3 = 3 and (0, 2) 2 = -(0) + 2 => 2 = 2]

Question 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

Answer:

Let y = work done

F = constant force = 5

x = distance travelled

Then y = 5x

For x = 0, y = 0, For x = 1, y = 5, For x = 2, y = 10

For x = 3, y = 15, For x = 4, y = 20

(i) Let A represent x = 2. The x-axis. From A, draw AB Ʇ x-axis,

Meeting the graph at B. From B, draw BC Ʇ y-axis, meeting y-axis

at C. The ordinate of C is 10.

So, when the distance travelled is 24 units, the work is done.

(ii) For x = 0, y = 0, the graph is shown above. Question 7:

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 x and Rs y.)

Draw the graph of the same.

Answer:

Let Yamini contributes X rupees and Fatima contributes Y rupees.

Then according to question,

X + Y  =100 ...........1

Put value of X = 0 in equation 1, we get

y = 100

Put value of X = 50 in equation 1, we get

50 + Y = 100

=> Y = 100 - 50

=> y = 50

Put value of X = 100 in equation 1, we get

100 + Y = 100

=> Y = 100 - 100

=> y = 0

So different solution of the equation 1 is:

X :  0       50      100

Y:  100     50       0

The graph is shown above. Question 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:

F = (9/5)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 300C, what is the temperature in Fahrenheit?

(iii) If the temperature is 950F, what is the temperature in Celsius?

(iv) If the temperature is 00C, what is the temperature in Fahrenheit and if the temperature is 00F, 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:

Given equation is

F = (9/5)C + 32 ....................1 (i) We have to take Celsius for x-axis and Fahrenheit for y-axis.

Now put C = 0 in equation 1, we get

F = (9/5)*0 + 32

=> F = 32

Again put C = 5 in equation 1, we get

F = (9/5)*5 + 32

=> F = 9 + 32

=> F = 41

Again put C = -5 in equation 1, we get

F = (9/5)*(-5) + 32

=> F = -9 + 32

=> F = 23

So

C:   0     5       -5

F:   32   41      23

(ii) When C = 300, the from equation 1

F = (9/5)*30 + 32

=> F = 9*6 + 32

=> F = 54 + 32

=> F = 86

(iii) When F = 950, the from equation 1

95 = (9/5)*C + 32

=> (9/5)*C = 95 - 32

=> (9/5)*C = 63

=> C = (63*5)/9

=> C = 7*5

=> C = 35

(iv) When C = 00, the from equation 1

F = (9/5)*0 + 32

=> F = 32

When F = 0, the from equation 1

0 = (9/5)*C + 32

=> (9/5)*C = - 32

=> C = (-32*5)/9

=> C = - 160/9

=> C = - 17.8

(v )Let X degree of celsius = X degree of Fahrenheit, then from equation 1, we get

X = (9/5)*X + 32

=> X - 32 = (9/5)*X

=> 5(X - 32) = 9X

=> 5X - 160 = 9X

=> 9X - 5X = -160

=> 4X = -160

=> X = -160/4

=> X = -40

So, at X = -40, both Celsius and Fahrenheit are same.

Exercise 4.4

Question 1:

Give the geometric representations of y = 3 as an equation

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

Answer:

Given, equation is: y = 3

(i) As an equation in one variable, it is the number 3 on the number line. (ii) As an equation in two variable, it can be written as 0 * x + y = 3

Value of x can be any number but y will continue to be 3.

It is a line parallel to x-axis and 3 units above it. Question 2:

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

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

Answer:

Given, equation is: 2x + 9 = 0

(i) As an equation in one variable.

2x + 9 = 0

=> 2x = -9

=> x = -9/2 (ii) The given equation can be written as:

2x + 0 * y + 9 = 0

This is an equation in two variables. Values of y can be any number but x remains -9/2.

It is a line parallel to y-axis and 9/2 units to the left of O. .