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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).

Solution:
Let the cost of a note book be Rs x and the cost of a pen be Rs y.
Then, according to the given condition of the question,
          x = 2y
⇒ x  2y = 0


 This is the required linear equation in two variables x and y.

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 - - 10 = 0

Solution:
(i) 2x + 3y = 9.35
⇒ 2x + 3y - 9.35 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = 3, c = -9.35.

(ii) x - -10 = 0
⇒ 5x – y - 50 = 0 (Multiplying both sides by 5)
Comparing with ax + by + c = 0, we get
a = 5, b = -1, c = -50

Question-3

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 = 6 (ii) x = 3y.

Solution:
(i)   2x + 3y = 6
    2x + 3y    6 = 0
Comparing with ax + by + c = 0, we get
a  2, b = 3, c = -6.
Another Method:
 
           2x + 3y = 6
⇒   2x + 3y    6 = 0
⇒ 2x  3y + 6 = 0
Comparing with ax + by + c = 0, we get (Multiplying both sides by  1)
a = 2, b  3, c = 6.

(ii) x = 3y
⇒ x   3y + 0 = 0
Comparing with ax + by + c = 0, we get
a = 1, b  3, c = 0

Question-4

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 =   5y (ii) 3x + 2 = 0

Solution:
(i)              2x 5y
                 2x =  5y
⇒ 2x + 5y + 0 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = 5, c = 0.

(ii)       3x + 2 = 0
            3x + 2 = 0
⇒  3x + 0y + 2 = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 0, c = 2.

Question-5

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) y   2 = 0 (ii) 5 = 2x.

Solution:
(i) y   2 = 0
Comparing with ax+by+c = 0
(0)x + (1)y + (  2) = 0
Therefore
a = 0, b = 1, c =  2

(ii) 5 = 2x
Comparing with ax + by + c = 0
2x    (0)y    5 = 0, we get
a = 2, b = 0, c =  5.

Question-6

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

Solution:
The option (iii) y = 3x + 5 has infinitely many solution is true.
Reason. 
For every value of x, there is a corresponding value of y and vice-versa.

Question-7

Write four solutions for each of the following equations:
(i) 2x + y = 7 (ii)  πx + y = 9

Solution:

(i) 2x + y = 7
    2x + y = 7
         ⇒y = 7 – 2x
Put x = 0, we get y = 7   2(0) = 7   0 = 7
Put x = 1, we get y = 7   2(1) = 7   2 = 5
Put x = 2, we get y = 7  2(2) = 7   4 = 3
Put x = 3, we get y = 7  2(3) = 7   6 = 1

∴ Four solutions are (0,7), (1, 5), (2, 3) and (3, 1).

(ii)   π x + y = 9
y = 9 - π x
Put x = 0, we get y = 9   π (0) = 9   0 = 9
Put x = 1, we get y = 9  π (1) = 9  π 
Put x =  1, we get y = 9   π (-1) = 9 +π 
Put x = , we get y = 9  π  = 9   9 = 0
∴ Four solutions are (0,9), (1, 9 -π ), (  1, 9 + π) and

Question-8

Write Four solutions for the following equation:  x = 4y

Solution:
x = 4y 
⇒ y =
Put x = 0, we get y == 0
Put x = 4, we get y == 1
Put x = - 4, we get y = = -1
Put x = 2, we get y ==
Four solutions are (0, 0), (4, 1), (- 4, - 1) and

Question-9

Check which of the following are solutions of the equation x   2y = 4 and which are not:
(i) (0, 2) (ii) (2, 0)

Solution:
The given equation is x  2y = 4 ...(1)
(i) (0, 2)
Put x = 0 and y = 2 in (1), we get
x - 2y = 0 - 2(2) =   4, which is not 4.
∴ (0, 2) is not a solution of (1).


(ii) (2, 0)
Put x = 2 and y = 0 in (1), we get
x  2y = 2   2(0) = 2  0 = 2,  which is not 4.
∴ (2, 0) is not a solution of (1).

Question-10

Check which of the following are solutions of the equation x  2y = 4 and which are not:
(i) (4, 0)
(ii) (
, 4) (iii) (1, 1).

Solution:
x  2y = 4 ------ (1)
(i) (4, 0)
Put x = 4 and y = 0 in (1), we get
x – 2y = 4  2(0) = 4
∴ (4, 0) is a solution of (1).


(ii) (, 4)
Put x =, y = 4 in (1), we get
x – 2y =     2(4) ⇒  8 7 which is not 4.
∴  (, 4 ) is not a solution of (1).


(iii) (1,1)
 Put x = 1 and y = 1 in (1), we get
x - 2y = 1   2(1) = 1   2 =   1, which is not 4.
∴ (1, 1) is not a solution of (1).

Question-11

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

Solution:
If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation So, putting x = 2 and y = 1 in the equation, we get
2(2) + 3(1) = k
 ⇒ 4 + 3 = k
 ⇒ k = 7
 
 

Question-12

Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4 (ii) x y = 2

Solution:
(i) x + y = 4
⇒ y = 4 - x
Table of solutions
 

x

1

2

y

3

2


We plot the points(1, 3) and (2, 2) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 4.

 

(ii) x – y = 2
     x – y = 2
 y = x - 2

Table of solutions

 
x 2 3
y 0 1


We plot the points(2, 0) and (3, 1) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x – y = 2.
 
 

Question-13

Draw the graph of the following linear equation in two variables: y = 3x

Solution:
y = 3x

Table of solutions  
x 0 1
y 0 3

 We plot the point (0, 0) and (1, 3) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 3x .

 

Question-14

Draw the graph of the following linear equation in two variables: 3 = 2x + y

Solution:
   3 = 2x + y

y = 3 – 2 x

Table of solutions
x 1 0
y 1 3

 We plot the points (1, 1) and (0, 3) on the graph paper and join the same by an ruler to get the line which is the graph of the equation 3 = 2 x + y.

 

Question-15

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

Solution:
Since the given solution is (2, 14)
Therefore, x = 2 and y = 14
Or, One equation is x + y = 2 + 14 = 16
x + y = 16
Second equation is x – y = 2 – 14 = -12
x - y = -12
Third equation is y = 7x
0 = 7x – y
7x - y = 0
In fact we can find infinite equations because through one point infinite lines pass.

Question-16

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

Solution:
 3y = ax + 7, then
 3(4) = a(3) + 7
⇒ 12 = 3a + 7
⇒ 3a = 12 – 7
⇒ 3a = 5
  ⇒ a =

Question-17

The taxi fare in a city is as follows: For the first kilometer, 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 far as Rs y, Write a linear equation for this information, and
draw its graph.

Solution:
Total distance covered = x km
Total fare = Rs y
Fair for the first kilometer = Rs.8
Subsequent distance = (x – 1) km
Fair for the subsequent distance = Rs 5(x – 1)

According to the question,
   y = 8 + 5(x – 1)
⇒ y = 8 + 5x – 5
y = 5x + 3

Table of solutions
 
x 0 1
y 3 8


We plot the points(0, 3) and (1, 8) on the graph paper and join the same by a ruler to get the line which is the graph of the equation
y = 5x + 3

Question-18

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

(i) y= x              (i) y = x + 2
(ii) x + y =0      (ii) y = x - 2
(iii) y = 2x         (iii) y = -x + 2
(iv) 2 + 3y =7x (iv) x + 2y = 6




Solution:
For Fig. (1). The Correct equation is (ii) x + y = 0
For Fig. (2). The Correct equation is (iii) y = -x + 2

Question-19

If the work done by a body on application of a constant force is directly proportional to the distance traveled 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 units

Solution:
Let the work done by the constant force by y units and the distance traveled by the body be x units.

Constant force = 5 units
We know that
Work done = Force ´ Displacement
          ⇒y = 5x

Table of solutions
 
x 0 1
y 0 5

We plot the points(0, 0) and (1, 5) on the graph paper and join the same by a ruler to get the line which is the graph of equation y = 5x.

 
From the graph,
Work done when the distance travelled is 2 units = 10 units
Work done when the distance travelled is 0 units = 0 units

Question-20

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.

Solution:
Let the contributions of Yamini and Fatima be Rs x and Rs y respectively.
Then according to the question

x + y = 100
This is the linear equation which the given data satisfies
Now, x + y = 100
         ⇒y = 100 – x

Table of solutions
 
x 0 50
y 100 50

We plot the points(0, 100) and (50, 50) on the graph paper and join the same by a ruler to get the line which is the graph of equation
x + y = 100.

 

Question-21

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 = 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 300 C, What is the temperature in Fahrenheit?

Solution:
(i) F =
We draw the X and Y axis with the each X co-ordinate = 10 units and each Y co-ordinate = 10 units



 

(ii) F =

When C = 30, then F =

                                 = 54 + 32

                                 = 86

 

∴ Required temperature = 860  F

Question-22

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 = C + 32

(i) If the temperature is 95°  F, What is the temperature in Celsius? 
 (ii) 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?

Solution:
(i) When F = 95, then
      F =

⇒95 =  

⇒  

⇒  

 ⇒ C =

  ⇒ C = 35

∴ Required temperature = 35° C

(ii) When C = 0, then F =

∴ Required temperature = 32°  F

Whan F = 0, then

    F =

⇒0 =



   ⇒ C =

         = -

∴ Required temperature = -° C

Question-23

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 = C + 32
Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it?

Solution:
 Let the temperature be x numerically. Then,

    ⇒ x = -

∴ Numerical value of required temperature = -40

Question-24

Give the geometric representations of y = 3 as an equation in one variable

Solution:
The given equation is y = 3
In one variable

The representation of y = 3 on the number line is as shown below:

 

Question-25

Give the geometric representations of y = 3 as an equation in two variables

Solution:
In two variables
y = 3
⇒ 0.x + 1.y = 3



 

It is a linear equation in two variables x and y. This is represented by a line. All the values of x are permissible because 0×x is always 0. However, y must satisfy the relation y = 3. Hence, two solutions of the given equation are x = 0, x = 2, y = 3


Thus the graph AB is a line parallel to the x-axis at a distance of 5 units above it.

Question-26

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

Solution:
The given equation is 2x + 9 = 0
In one variable
2x + 9 = 0 ⇒ 2x = -9 ⇒ x = -

The representation of 2x +9 on the number line is as shown below:


 

Question-27

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

Solution:
In two variables
2x + 9 = 0 ⇒ 2x + 0y + 9 = 0

It is a linear equation in two variables x and y. This is represented by a line. All the values of y are permissible because 0y is always 0. However, x must satisfy the relation 2x + 9 = 0, i.e. x = -, y = 2

The graph AB is a line parallel to the y-axis and at a distance units to the left of it.

 





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