Class 9 Maths Chapter 4 Linear Equations in Two 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)
Solution:
Let the cost of a notebook =Rs.x
and the cost of a pen = Rs.
y According to question,
Cost of a notebook = 2 (cost of a pen)
x = 2y ⇒ x -2y = 0

 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 = 9.3 
(ii) x-  – 10 = 0
(iii) – 2x + 3y = 6
(iv) x = 3y
(v) 2x = – 5y
(vi) 3x+2=0
(vii) y-2=0
(viii) 5=2x
Solution:
(i) Given linear equation, 2x + 3y= 9.35
∴ (2)x + (3)y + (- 9.3 ) = 0
On comparing with ax ‘+ bx + c = 0, we have
a = 2, b = 3 and c = – 9.3

(ii) Given linear equation x--10 =0
⇒ x+(-) y+(-10)=0
On comparing with ax + by + c = 0, we get
a = 1, b = – and c = -10

(iii) Given linear equation -2x + 3y = 6
⇒ -2x + 3 y – 6 = 0
⇒ (-2)x + (3) y + (-6) = 0
On Comparing with ax + by + c – 0, we get
a = – 2, b = 3 and c = – 6

(iv) Given linear equation x = 3y
⇒ x-3y=0
⇒(1) x + (-3) y + 0 = 0
On comparing with ax + by + c = 0, we get
a = 1, b = -3 and c = 0.

(v) Given linear equation 2x = – 5y
⇒ 2x + 5y = 0
⇒ (2)x + (5)y + 0 = 0
On comparing with ax + by + c = 0, we get
a – 2, b = 5 and c = 0.

(vi) Given linear equation 3x + 2 = 0
⇒ 3x +2 + 0y = 0
⇒  (3)x + (0)y + (2) = 0
On comparing with ax + by + c = 0 , we get
a = 3, b = 0 and c = 2.

(vii) Given linear equation y – 2=0
⇒ (0)x + (l)y + (-2) = 0
On comparing with ax + by + c = 0, we get
a= 0, b-1 and c = – 2

(viii) Given linear equation
5= 2x
⇒ 5 – 2x = 0
⇒  -2x+0y + 5 = 0
⇒ (-2)x + (0)y + (5) = 0
On comparing with ax + by-c = 0, we get
a= – 2, b = 0 and 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 ,
Solution:
(iii) We know that, a linear equation in two variables has infinitely many solutions. Here, given equation is a linear equation in two variables x and y, so it has infinitely many solutions.

 Question 2.
Write four solutions for each of the following equations :
(i) 2x + y = 7,
(ii) πx + y = 9
(iii) x = 4y Sol.
Solution:
(i) Given equation, 2x + y=7        …(i)
Now, putting x = 0, in eq. (i), we get 2(0) + y=7
⇒ 0 + y = 7 => y=7
So, x = 0 and y = 7 is a solution of given equation.
On putting x = 1 in eq (i), we get
2(1) + y =7 ⇒ y=7-2    ⇒ y=5
So, x = 1 and y = 5 is a solution of given equation
On putting x = 2, in eq. (i) we get, 2(2) + y = 7 ⇒ y= 7 – 4 ⇒ y= 3
So, x = 2 and y = 3 is a solution of given equation.
On putting x = 3 in eq(i), we get 2(3) + y =7
⇒ y= 7-6 ⇒  y = 1
So, x = 3 and y = 1 is a solution of given equation.
Hence, four out of the infinitely many solutions of the given equation are (0, 7), (1, 5) (2, 3) and (3,1).

(ii) Given equation πx + y =9 – – – (i)
On putting x = 0 in eq. (i) we get, π(0) + y=9
⇒ y = 9 – 0 =>y = 9
So, x = 0 and y = 9 is a solution of given equation.
On putting x = 1 in eq. (i), we get π(1) + y= 9
y-9-π
So, x = 1, and y = (9 -π) is a solution of given equation.
On putting x = 2 in eq (i), we get π(2) + y=9
y = 9 – 2π
So x = 2 and y = (9 – 2π) is a solution of given equation.
On putting x = -1 in eq (i), we get π(-9) + y=9
y = 9 + π
So x = -1 and y = (9 + π) is a solution of given equation.
Hence, four out of the infinitely many solutions of the given equation are
(0, 9), {1,(9 – π)}, {2,(9 – 2π)}, {-1, (9 + π)}

(iii) Given equation x = 4y     …(i)
On putting x = 0, in eq. (i), we get 4y=0 ⇒ y=0
So, x – 0 and y = 0 is a solution of given equation.
On putting x = 1 in eq. (i), we get 4y=1 ⇒ y= 
So, x -1 and y =  is a solution of given equation.
On putting x – 4, in eq (i), we get
4y= 4
⇒ y= –  =>y = 1
So, x = 4 and y = 1 is a solution of given equation, on putting x = – 4 in eq. (i), we get
4y= -4  ⇒ y=  = -1
So, x = – 4 and y = -1 is a solution of given equation.
Hence, four out of the infinitely many solutions of given equation are
(0, 0), (1, ), (4,1) and (-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) ( ,4 )
(v) (1, 1)                                                            .
Solution:
(i) Given equation is x – 2y = 4.
(0, 2) means x = 0 and y = 2
On putting x = 0 and y = 2 in L.H.S., get
L.H.S. = x – 2y = 0 – 2(2) = – 4
But     R.H.S. = 4
L.H.S ≠  R.H.S
Hence (0, 2) is not a solution of x – 2y – 4.

(ii) Given equation is x – 2y = 4 (2, 0) means x – 2 and y = 0
On putting x = 2 and y = 0 in L.H.S., we get
L.H.S. = x – 2y = 2 – 2(0) = 2-0 = 2
But – R.H.S. = 4
∴ L.H.S = R.H.S.
Hence (2, 0) is not a solution of x – 2y – 4.

(iii) Given equation is x – 2y = 4 (4, 0) means x = 4 and y = 0.
On putting x = 4 and y = 0 in L.H.S., we get
L.H.S. = x – 2y = 4 – 2(0) =4-0 = 4
But         R.H.S.  = 4
∴ L.H.S. = R.H.S.
Hence, (4, 0) is a solution of x – 2y = 4.

(iv) Given equation is x – 2y = 4
(, 4) means x =  and y = 4
On putting x =  and y = 4 in L.H.S., we get
L.H.S. = x – 2y =  – 2(4) = -8
= (1 – 8) = – 7
But  R.H.S.  = 4
∴ L.H.S ≠ R.H.S.
Hence (, 4) is not a solution of x – 2y = 4

(v) Given equation is x – 2y = 4
(1, 1) means x = 1 and y = 1
On putting x = 1 and y = 1 in L.H.S. we get
L.H.S. = x – 2y = 1 – 2(1) = 1 – 2 = -1 But        R.H.S.  = 4
∴  L.H.S. ≠ R.H.S.
Hence, (1, 1) is not a solution of x – 2y = 4.

 Question 4.
Find the value by k if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
If x = 2, y = 1 is a solution by the equation 2x + 3y = k, then these value will satisfy the equation.
∴ 2 x 2+3 x 1 = k ⇒ k=4+3=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
Solution:
(i) Given equation is x + y – 4
⇒ y= 4 – x
To draw the graph, we need atleast three solutions of the equation When x = 0, then y= 4 – 0 = 4; When x = 1, then y= 4-1 = 3 When x = 2, then y – 4-2 = 2
Thus, we have the following table

X	0	1	2
y	4	3	2

Plot the points (0, 4), (1, 3) and (2, 2) on the graph paper and join them we get a line AB as shown below :

Thus, the line AB is the required graph of the given linear equation,

(ii) Given equation is x-y = 2 => y = x – 2
To draw the graph, we need atleast three solutions of the equation. When x = 0, then y = 0 – 2 = -2; When x = 1, then y = 1- 2 = -1 When x = 2, then y = 2 – 2 = 0

Then we have the following table :

X	0	1	2
y	-2	– 1	0

So, plot the points (0, – 2), (1, -1) and (2, 0) on the graph paper. Joining these points, we get a straight line PQ.

Thus, the line PQ is required graph of the given equation.

(iii) Given equation is y = 3x
To draw the graph, we need at least three solutions to the question.
When x – 0, then y = 3(0) ⇒ y = 0
When x = 1, then y = 3(1) ⇒ y = 3
When x = -1, then y = 3(-1) ⇒ y = – 3

Then, we have the following table :

X	0	1	– 1
y	0	3	-3

So, plot the points (0, 0), (1, 3) and (-1, -3) on the graph paper. Joining these points, we get a straight line LM

Thus, LM is the required graph of the given equation.
Note: The graph of the equation of the form y – kx is a straight line which always passes through the origin.

(iv) Given equation is 3 = 2x + y .
=> y – 3 – 2x
To draw the graph, we need at least three solutions to the equation
When x = 0 then y = 3 – 2 (0) ⇒ y = 3
When x = 1, then y = 3 – 2(1) ⇒ y = 1
When x = 2, then y = 3 – 2 (2) ⇒ y = – 1

Then, we have the following table :

X	0	1	2
y	3	1	– 1

So, plot the points (0, 3), (1, 1) and (2, -1) on the graph paper. Joining these points, we get a straight line CD.

Thus, the line CD is the required graph of the given equation

 Question 2.
Give the equations of two lines passing through (2, 14), How many more such lines are there, and why?
Solution:
Let  7x – y = 0  …. (i)
2x + y=18
are two linear equations.
We observe equations (i) and (ii) by  substituting x = 2 and y = 14.
From eq. (i), we have
7 (2) -14 = 14 -14 = 0
i.e., (2, 14) is the solution of eq. (i).
From equation (ii), we have 2(2) +14 = 4 +14 = 18 It is quite obvious that graphs of equations (i) and (ii) are two straight lines and the point (2,14) lies on both the lines. Hence, the two lines pass through the point (2, 14).
We can find infinitely many linear equations whose one pair of the solution is (2, 14) and correspondingly, there are infinitely many straight lines in the Cartesian plane which pass through the point (2, 14).

 Question 3.
If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
If the point (3, 4) lies on the graph, then it will satisfy the equations.
Hence, 3(4) – a (3)- 7 = 0
3 (4) = a (3) + 7 ⇒ 12 = 3a+7
⇒ 12 – 7 = 3a    ⇒ 3a = 5
a = 

 Question 4.
The taxi fare in a city is as follows:
For the first kilometer, 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:
Given, total distance covered = x km
Total taxi fare = Rs. y
Fare for the first kilometer =Rs. 8
Fare for subsequent distance =Rs. 5 per km
Remaining distance = (x -1) km Fare for next (x -1) km = Rs. 5 x (x -1)
Total taxi fare = Rs. 8 + Rs. 5(x -1)
According to the question,
⇒  y = 8 + 5 (x-1)
⇒  y = 8 + 5x – 5
⇒  5x – y + 3 = 0
Which is the required linear equation.
It can also be written as y = 5x + 3
When x = 0,   then y =  5 (0) + 3            ⇒ y = 3
When x = -1,   then   y= 5 (-1) + 3          ⇒  y=-2
When x = – 2,  then   y =   5 (-2) + 3      ⇒  y = – 7
Thus, we get the  following table :

Now, plotting the points (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

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


Solution:
In figure (i), given points are (0, 0), (-1,1) and (1, -1)
So, these points will satisfy the equation of the line.
At (0, 0), x + y – 0 + 0 = 0
At (-1,1),    x + y = -1+1=0
At (1,-1), x + y = 1-1=0
i.e., these points satisfy the equation x + y = 0.
Hence, given graph represents the equation x + y = 0.
In figure (ii), given points are (-1, 3), (0, 2) and (2, 0).
So, these points will satisfy the equation of line
At (-1, 3), x + y = -1 + 3 = 2
At (0, 2), x + y = 0 + 2 = 2
At (2, 0), x + y = 2 + 0 = 2
i.e., these points satisfy the equation x + y = 2.
Hence, given graph represents the equation x + y = 2.
i.e.,      y = – x + 2

 Question 6.
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 traveled by the body is
(i) 2 units
(ii) 0 unit
Solution:
Given, work done by a body on application of a constant force is directly proportional to the distance traveled by the body. i.e., Work done (w)a distance (s)
⇒ W = F.S.
Where, F is an arbitrary constant
Now, let x be the distance and y be the work done. Then,
y=Fx
Which is the required linear equation in two variable x and y. If the constant force is 5 units i.e., F = 5, then the equation is 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
When   x = 2, then y = 5 (2) = 10
Thus, we have the following table :

X	0	1	1.5	2
y	0	5	7.5	10

Now, plotting the points (0, 0), (1, 5) and (1.5, 7.5) on 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 = 5(2) = 10 units.
Hence, the work done= 10 units.

(ii) Distance travelled = 0 unit.
 ∴ y = 5(0) = 0
Hence, the work done = 0 units.

 Question 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 Rs. x, and Rs. y). Draw the graph of the same.
Solution:
Let the contributions of Yamini and Fatima towards the Prime Minister’s Relief Fund to help the earthquake victims be Rs. x and Rs. y respectively. Then, According to the question, x + y = 100
=> y = 100 – x
When x = 0, then y=100 – 0 =100 When x = 50, then y = 100 – 50 = 50 When x = 100, then
y = 100 -100 = 0
We get the following table :

X	0	50	100
y	100	50	0

For drawing the graph, plot the ordered pairs (0,100), (50, 50) and (100, 0)
on a graph paper .Joining these point we get a line PQ.

Thus, PQ is the required graph of x + y = 100

 Question 8.
In countries like the USA and Canada, a 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 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) 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?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Solution:
(i) we have F =  C+32 ………….(i)
when C = 0,then  F=  x 0+32 = 32
when C = -15,then F=  (-15) + 32 =- 27 + 32 = 5
when C = – 10,  then F=  (-10) + 32=9 (-2) + 32 = 14

We have the following table :

c	0	-15	– 10
F	32	5	14

Let us take Celsius along x-axis and Fahrenheit along y-axis and plot the points (0, 32), (-15,5) and (-10,14) on a graph paper using suitable scale (1 cm = 10 units on x-axis and 1 cm =10 units on y-axis)

(ii) From the graph, we have 86°F corresponds to 30°C
(iii) From the graph, we have  95°F corresponds to 35°C
(iv) From the graph, we have 32°F corresponds to 0°C
And 17.8°C corresponds to 0°F.
(v) Yes, there is a temperature which is numerically the same in both Fahrenheit and Celsius.
On putting C = F in F =  C+32, we get
F= F +32
= 5F=9F+160
= 9F – 5F =-160
=  4F=-160
F = -40
Thus, 400 F is same as – 40°C.

 

Exercise 4.4

 Question 1.
Give the geometric representations of y – 3 as an equation
(i) in one variable.
(ii) in two variables.
Solution:
The given linear equation is
y = 3 …(i)
(i) The representation of the solution on the number line is shown in the figure below, where y = 3 is treated as an equation in one variable.

(ii) We know that y = 3 can be written as
0. x + y = 3
Which is a linear equation in the variables x and y. This is represented by a line. Now, all the values of x are permissible because 0.x is always 0. However, y must satisfy the equation y = 3.
Note that, the graph AB is a line parallel to the x-axis and at a distance of 3 units of the upper side of it.

 Question 2.
Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables
Solution:
Given linear equation is 2x + 9 = 0 i.e., x = – 
(i) If x = –  is treated as an equation in one variable, then it has a unique
solution x = – .
So, it is a point on the number line as shown below :

(ii) Given equation 2x + 9 = 0 can be written as 2x + 0 . y + 9 = 0, which is a linear equation in two variables x and y.
We can write the given equation as,

When y=1, then  = – 
When y=2, then  = – 
When y=3, then  = – 

X	–	–	–
y	1	2	3

Now Playing the points – , –  –  paper and joining then ,we get a line PQ as a solution of 2x+9=0

Thus, the graph PQ is a line parallel to the y-axis at a distance of units in the direction of negative X-axis.