Exercise 4.1

Q.1       The cost of a notebook is twice the cost of a pen. Write  a linear equation in two  variables to represent this statement.
Sol. 

Let the cost of notebook be  Rs x and that of a pen  be Rs y.
Since  the cost of a notebook is twice  the cost  of pen. So, the  required linear equation in two
variables to represent  the above  statement is given by
x = 2y.
⇒ x – 2y = 0

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Q.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−y5−10=0
                (iii) – 2x + 3y = 6                                    
             (iv) x = 3y
                (v) 2x = – 5y     
             (vi) 3x + 2 = 0
                (vii) y – 2 = 0                                         
             (viii)  5 = 2x
Sol.

(i)  2x+3y=9 can be written as 2x+3y−9.35¯¯¯=0
On comparing  it with ax + by + c = 0 , we have
a = 2, b = 3 and  c=−9.35¯¯¯
(ii)  On comparing  x−y5−10=0  with  ax  + by + c = 0, we have,
a=1,b=−15andc=−10
(iii)  –2x + 3y = 6 can be  written as – 2x + 3y – 6 = 0
On  comparing  it with  ax + by + c = 0 , we have
a = – 2, b = 3 and c = – 6
(iv)  x = 3y can  be written  as x – 3y = 0
On  comparing  it with  ax + by + c = 0 , we have
a = 1, b = – 3 and c = 0
(v)  2x = – 5y can be written  as  2x + 5y = 0
On  comparing  it with  ax + by + c = 0 , we have
a = 2, b = 5 and c = 0
(vi)  On  comparing  3x + 2 = 0 with  ax + by + c = 0 , we  have
a = 3, b = 0 and c = 2
(vii) On  comparing  y – 2 = 0 with  ax + by + c = 0,  we  have
a = 0 , b = 1 and c = – 2.
(viii)  5 = 2x can be written  as 2x – 5 = 0
On  comparing  with  ax + by + c = 0 , we have
a = 2, b = 0  and c = –5.
or    – 2x + 5 = 0 , we have  a = – 2, b = 0 , c  = 5

Exercise 4.2

Q.1      Which one of the following options is true, and why?
                y = 3x + 5 has
                (i) a unique solution
                (ii) only two solution
                (iii) infinitely many solutions
Sol.

Given equation is y = 3 x + 5
For y = 0 , 3x + 5 = 0 ⇒x=−53
Therefore (−53,0) is one of solution.
For x = 0, y = 0 + 5 = 5
Therefore, (0, 5) is another solution.
For x = 1, y = 3 × 1 + 5 = 8
Therefore (1, 8) is another solution.
Clearly, for different values of x, we get another value of y. Thus, the chosen value of x together with
this value of y constitutes another solution of the given equation. So, there is no end to different
solutions of a linear equation in two variables.
Therefore, A linear equation in two variables has infinitely many solutions.

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Q.2      Write four solutions for each of the following equations :
                (i) 2x + y = 7                
             (ii) πx+y=9                          
             (iii) x = 4y
Sol.

(i) The given equation can be written as y = 7 – 2x.
For x = 0 , y = 7 – 2 × 0 = 7 – 0 = 7
For x = 1, y = 7 – 2 × 1 = 7 – 2 = 5
For x = 2, y = 7 – 2 × 2 = 7 – 4 = 3
For x = 3, y = 7 – 2 × 3 = 7 – 6 = 1
Therefore the four solutions of the given equation are (0, 7), (1, 5) , (2, 3) and (3, 1) .
(ii) The given equation can be written as  y=9−πx
For x = 0, y = 9 – 0 = 9
For x = 1, y=9−π
For x = 2 , y=9−2π
For x = 3, y=9−3π
Therefore the four solutions of the given equation are (0, 9),
(1,9−π) , (2,9 −2π) and (3, 9 −3π)
(iii) The given equation can be written as x = 4y
For x = 0,   y = 0
For  x = 1 ;  y=14
For x = 2 ;  y=24=12
For x = 3 ; y=34=34
Therefore the four solutions of the given equation are (0,0)(1,14)(2,12)and(3,34)

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Q.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–√,42–√)(v) (1, 1)
Sol.

(i) Putting x = 0 , y = 2 in L.H.S. of x – 2y = 4, we have
L.H.S. = 0 – 2 × 2 = – 4 ≠ R.H.S.
Therefore x = 0 , y = 2 is not its solution.
(ii) Putting x = 2, y = 0 in L.H.S. of x – 2y = 4, we have
L.H.S. = 2 – 2 × 0 = 2 – 0 = 2 ≠ R.H.S.
Therefore x = 2, y = 0 is not its solution.
(iii) Putting x = 4, y = 0 in the L.H.S. of x – 2y = 4, we have
L.H.S. = 4 – 0 = 4 = R.H.S.
Therefore x = 4, y = 0 is its solution.
(iv) Putting x=2–√,y=42–√ in the L.H.S. of x – 2y = 4, we have
L.H.S. x=2–√−2×42–√=2–√−82–√
=−72–√≠ R.H.S.
Therefore (2–√,42–√) is not its solution.
(v) Putting x = 1, y = 1 in the L.H.S. of x – 2y = 4, we have
L.H.S. = 1 – 2 × 1 = 1 – 2 = – 1 ≠ R.H.S.
Therefore x = 1, y = 1 is not its solution.

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

If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation.
Therefore 2 × 2 + 3 × 1 = k ⇒ k = 4 + 3 = 7.

Exercise 4.3

Q.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) We have x + y = 4 ⇒ y = 4 – x
When x = 0 , we have : y = 4 – 0 = 4
When x = 2, we have : y = 4 – 2 = 2
When x = 4, we have : y = 4 – 4 = 0
Thus, we have the following table :

Plotting the points (0, 4) (2, 2) and (4, 0) on the graph paper and drawing a line joining them, we obtain
the graph of the line represented by the given equation as shown.

(ii) We have, x – y = 2 ⇒ y = x – 2
When x = 0 , we have y = 0 – 2 = – 2
When x = 2, we have y = 2 – 2 = 0
When x = 4, we have y = 4 – 2 = 2
Thus, we have the following table :

Plotting the points (0, –2), (2, 0) and (4, 2) on the graph paper and drawing a line joining them, we obtain
the graph of the line represented by the given equation as shown.

(iii)We have y = 3x
When x = 0 , we have : y = 3(0) = 0
When x = 1, we have : y = 3 (1) = 3
When x = –1, we have : y = 3(–1) = – 3
Thus , we have the following table :

Plotting the points (0, 0), (1, 3) and (–1, –3) on the graph paper and drawing a line joining them, we
obtain the graph of the line represented by the given equation as shown.
(iv) We have , 3 = 2x + y ⇒ y = 3 – 2x
When x = 0, we have :
y = 3 – 2(0) = 3 – 0 = 3
When x = 3, we have :
y = 3 – 2(3) = 3 – 6 = – 3
When x = – 1, we have :
y = 3 – 2 (–1) = 3 + 2 = 5
Thus , we have the following table :

Plotting the points (0, 3), (3, –3) and (–1, 5) on the graph paper and drawing a line joining them, we
obtain the graph of the line represented by the given equation as shown

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

Here (2, 14) is a solution of a linear equation we are looking for. One example of such a linear equation is 7x – y = 0.  Note that  x + y = 16, 2x + y = 18 and 7x + y = 28 are also satisfied by the co-ordinates of the point (2, 14). So any line passing through the point (2, 14) is an example of a linear equation for which (2, 14) is a solution. Thus there are infinite number of lines through (2, 14).

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

Since (3, 4) lies on the graph corresponding to 3y = ax + 7. Therefore, (3, 4) satisfies the given equation.
That is , 3(4) = a(3) + 7
⇒ 12 – 7 = 3a
⇒ 3a = 5
⇒ a=53
Hence for a=53, (3, 4) lie on the graph of equation 3y = ax + 7.

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Q.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 Rs y, write a linear equation for this information and draw its graph.
Sol.

Taxi fare for first km = Rs 8
Taxi fare for the subsequent km = Rs 5
Total fare = Rs y
Total distance = x km
The linear equation for the above information is given by
y = 8 × 1 + 5(x – 1)
⇒ y = 8 + 5x – 5
⇒ y = 5x + 3
⇒ 5x – y + 3 = 0
When x = 0, y = 5 × 0 + 3 = 0 + 3 = 3
When x = – 1, y = 5 ×( – 1) + 3 = – 5 + 3 = – 2
When x = – 2, y = 5 × (– 2) + 3 = – 10 + 3 = – 7
Thus, we have the following table :

Plotting the points (0, 3), (–1, –2) and (–2, –7) on the graph paper and drawing a line joining them, we,obtain the required graph.

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Q.5     From the choices given below, choose the equation whose graphs are given in Fig(a) and Fig(b)
           For figure (a)                 For figure (b)
              (i) y = x b                        (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

                  

Sol.

For Fig (a), the correct equation from the choices given is clearly x + y = 0 as it is satisified by the points (–1,1) and (1, –1) given on the graph.
For Fig (b), the correct equation from the choices given is clearly y = – x + 2 as it is satisfied by the points (–1, 3) (0, 2) and (2, 0) given on the graph.

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Q.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.

Let x be the distance and y be the work done. Therefore, according to the problem the equation will be y = 5x.
To draw its graph :
When x = 0 , we have , y = 5(0) = 0
When x = 1, we have y = 5 (1) = 5
When x = – 1, we have y = 5 (–1) = – 5
Therefore the table is as under :

Plotting (0, 0) , (1,5) and (–1, –5) on the graph paper and drawing a line joining them we obtain the graph of
the line y = 5x as shown.
From the graph , clearly,
(i) when distance travelled is 2 units. i.e. x = 2, then y = 10.
Therefore Work done = 10
(ii) When distance travelled is 0 units i.e., x = 0 , then y = 0
Therefore Work done = 0.

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Q.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.
Sol.

Let Yamini and Fatima contributed Rs x and Rs y towards the P.M.’s Relief Fund totally Rs 100.
Therefore the linear equation using the above data is x + y = 100, i.e. y = 100 – x.
To draw its graph :
When x = 0, we have y = 100 – 0 = 100
When x = 100 we have y = 100 – 100 = 0
When x = 50 , we have y = 100 – 50 = 50
The table for these values is as under

Plotting the points (0, 100), (100, 0) and (50, 50) on the graph paper and drawing a line joining them, we obtain the graph of the line x + y = 100 as shown.

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Q.8     In countries like the 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=(95)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 30ºC, what is the temperature in Fahrenheit ?
(iii) If the temperature is 95º F, what is the temperature in Celsius ?
(iv) If the temperature is 0º C, what is the temperature in Fahrenheit and if the temperature is 0ºF, 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.
Sol.

(i) We have F=95C+32⇒C=59(F−32)
We calculate the vales of F for different values of C.
When C=−40,F=95×(−40)+32=−72+32=−40
When C=10,F=95×10+32=18+32=50
The table of values is as under :

We choose a suitable scale on the x-axis (for Celsius) and on the same scale on the y-axis (for Fahrenheit).
Plotting the points (– 40, – 40) and (10, 50) on the graph point. On joining these points by line segment, we obtain the graph of  F=95C+32.
(ii) From the graph, we see that when C = 30º shown C1 on the x-axis in the positive direction, then F =86º shown by the point F1 on the y-axis in the positive direction.
Hence 30ºC = 86ºF
(iii) From the graph, we see that when F = 95º shown F2 on the y-axis in the positive direction, then C = 35º shown by the point C2 on the x-axis in the positive direction.
Hence 95º F = 35ºC
(iv) From the graph, clearly 0ºC = 32ºF and 0ºF = – 17.8ºF.
(v) Clearly from the graph the temperature which is numerically the same in both Fahrenheit and Celsius is – 40º i.e. – 40ºC = – 40º F.

Exercise 4.4

Q.1      Give the geometric representation of y = 3 as an equation.
                (i) in one variable           (ii) in two variables
Sol.

(i) The representation of the solution on the  number when y = 3 is treated as an equation in one variable is as under

(ii)  We know that y = 3 can be written as 0. x + y = 3 as a linear equation in variables x and y. Now all the values of x are permissible as 0. x is always 0. However, y must satisfy the  relation y = 3.
Hence , three  solution of the given equation are
x = 0, y = 3 ;      x = 2, y = 3   ;    x = – 2 , y = 3

Plotting the  points (0, 3) (2, 3) and  (–2, 3) and on joining them we get the graph AB as a line  parallel to x-axis at a distance of 3 units above it.

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Q.2       Given the geometric representations of 2x + 9 = 0  as an equation.
                 (i)  in one  variable             (ii)  in two variables
Sol.

(i) The  representation of the solution on the number line 2x + 9 = 0 i.e., x=−92 is treated as an  equation in one variable is as under. (ii) We know  that 2x + 9 = 0 can be written as 2x + 0y + 9  = 0 as a linear equation in variables x and y. Now all the  values of x are  permissible  as 0. y is always 0. However, x must  satisfy the relation 2x + 9 = 0
i.e., x=−92.
Hence three solution of the  given  equation are  x=−92,y = 0 ; x=−92,y = 2 and x=−92,y =– 2.
Therefore  Plotting  the point (−92,0),(−92,2)and(−92,−2)and on  joining them we get the graph AB as a line  parallel to y-axis at a distance of 92 on the left of y-axis.