NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables

Q1 ) 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 :

The iii) option is true.

A linear equation in two variables has infinitely many solutions.

Q2 ) Write four solutions for each of the following equations :
i) 2x + y = 7
ii) \(\pi\)x + y = 9
iii) x = 4y

Answer :

i)By inspection, x = 2 and y = 3 is a solution because for these values of x and y,

we get,2(2) + 3 = 4 + 3 = 7

Now, if we put x = 0, then the equation reduces to an unique solution of y = 7.

So, one solution of 2x + y = 7 is (0, 7).

Similarly, if we put y = 0, then the equation reduces to an unique solution of x = \(\frac{7}{2} \).

So, another solution of 2x + y = 7 will be (\( \frac{7}{2} \) , 0).

Finally, let us put x = 1, then,

we get, 2(1) + y = 7
=> 2 + y = 7
=> y = 5

Hence, (1, 5) is also a solution of given equation.

Therefore, four of the infinitely many solutions of the equation 2x + y = 7 are :
(2, 3) , (0, 7) , (\(\frac{7}{2} \) , 0) and (1, 5).


ii)By inspection, x = \(\frac{1}{\pi } \) and y = 8 is a solution because for these values of x and y,

we get,\(\pi (\frac{1}{\pi} \)) + 8 = 1 + 8 = 9

Now, if we put x = 0, then the equation reduces to an unique solution of y = 9.

So, one solution of \(\pi\)x + y = 9 is (0, 9)).

Similarly, if we put y = 0, then the equation reduces to an unique solution of x = \(\frac{9}{\pi } \).

So, another solution of 2x + y = 7 will be (\(\frac{9}{\pi } \), 0).

Finally, let us put x = 1, then,

we get, \(\pi\)(1) + y = 9
=> \(\pi\) + y = 9
=> y = 9 - \(\pi\)

Hence, (1, 9 - \(\pi\)) is also a solution of given equation.

Therefore, four of the infinitely many solutions of the equation 2x + y = 7 are :
(\(\frac{1}{ \pi } \), 8) , (0, 9) , (\(\frac{9}{\pi }\), 0) and (1, 9 - \(\pi\)).


iii)By inspection, x = 0 and y = 0 is a solution because for these values of x and y,

we get,0 = 4(0) => 0 = 0
Now, if we put x = 4, then the equation reduces to solution of y = 1.

So, one solution of x = 4y is (4, 1).

Similarly, if we put y = -1, then the equation reduces to solution of x = -4.

So, another solution of x = 4y will be (-4, -1).

Finally, let us put y = \(\frac{1}{2} \) , then,
we get, x = 4(\(\frac{1}{2} \))
=> x = 2

Hence, (2, \(\frac{1}{2} \) ) is also a solution of given equation.

Therefore, four of the infinitely many solutions of the equation x = 4y are :
(0, 0) , (4, 1) , (-4, -1) and (2, \(\frac{1}{2} \) ).

Q3 ) Check of the following are solution of the equation x – 2y = 4 and which are not?
i) (0, 2)
ii) (2, 0)
iii) (4, 0)
iv) (\(\sqrt{2}\), 4\(\sqrt{2}\))
v) (1, 1)

Answer :

If (x, y) are the solutions of the equation x – 2y = 4 then, they satisfies that equation.

i) Let's put x = 0 and y = 2,
we get, L.H.S. = (0) - 2(2) = -4
so, since, L.H.S. \(\ne\) R.H.S.
(0, 2) are not the solutions of given equation.

ii) Let's put x = 2 and y = 0,
we get, L.H.S. = (2) - 2(0) = 2
so, since, L.H.S. \(\ne\) R.H.S.
(2, 0) are not the solutions of given equation.

iii) Let's put x = 4 and y = 0,
we get, L.H.S. = (4) - 2(0) = 4
so, since, L.H.S. = R.H.S.
(4, 0) are the solutions of given equation.

iv) Let's put x = \(\sqrt{2}\) and y = 4\(\sqrt{2}\),
we get, L.H.S. = (\(\sqrt{2}\)) - 2(4\(\sqrt{2}\)) = -7\(\sqrt{2}\)
so, since, L.H.S. \(\ne\) R.H.S.
(\(\sqrt{2}\), 4\(\sqrt{2}\)) are not the solutions of given equation.

v) Let's put x = 1 and y = 1,
we get, L.H.S. = (1) - 2(1) = -1
so, since, L.H.S. \(\ne\) R.H.S.
(1, 1) are not the solutions of given equation.

Q4 ) Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Answer :

As x = 2 and y = 1 is the solution of the equation 2x + 3y = k, so it satisfies the equation.

put x = 2 and y = 1,
we get,
k = 2(2) + 3(1)
= 4 + 3
= 7

Therefore, the value of k is 7.

Q1 ) 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) To draw the graph, we need atleast two solutions of the equation.

x	0	4
y	4	0
(x, y)	(0, 4)	(4, 0)

Thus, we draw the graph by plotting the two points from table and then joining by a line.

ii) To draw the graph, we need atleast two solutions of the equation.

x	0	2
y	-2	0
(x, y)	(0, -2)	(2, 0)

Thus, we draw the graph by plotting the two points from table and then joining by a line.


iii)To draw the graph, we need atleast two solutions of the equation.

x	0	1	-1
y	0	3	-3
(x, y)	(0, -2)	(2, 0)	(-1, -3)

Thus, we draw the graph by plotting the two points from table and then joining by a line.


iv)To draw the graph, we need atleast two solutions of the equation.

x	0	1	-1
y	3	1	5
(x, y)	(0, 3)	(1, 1)	(-1, 5)

Thus, we draw the graph by plotting the two points from table and then joining by a line.

Q2 ) Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer :

Here, (2, 14) is a solution of a linear equation.

One example of such a linear equation is y = 6x + 2 and another is x + y = 16.

There are infinitely many lines because there are infinitely many linear equations which are satisfied by the coordinates of the point (2, 14).

Q3 ) If the point (3, 4) lies on the graph of the equation 3y - ax - 7 , find the value of a .

Answer :

If the point (3, 4) lies on the graph, then it satisfies the equation.

\(\because \) 3y - ax - 7 = 0
\(\Rightarrow \) 3(4) - a(3) - 7 = 0
\(\Rightarrow \) 12 - 3a = 7
\(\Rightarrow \) 12 - 7 = 3a
\(\Rightarrow \) 5 = 3a
\(\Rightarrow \) a = \(\frac{5}{3} \)

Q4 ) 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 the total fare as Rs. y, write a linear equation for this information and draw its graph.

Answer :

Distance covered = x km = 1 + (x – 1) km

The fare for the first kilometer is Rs. 8.

Fare for next (x - 1)km
= (x - 1) × 5
= 5(x - 1).

According to question, Total fare = y.
\(\therefore \) 8 + 5(x - 1) = y
\(\Rightarrow \) 8 + 5x - 5 = y

So, the equation becomes 5x - y + 3 = 0

x	0	1
y	3	8
(x, y)	(0, 3)	(1, 8)

Now, plot the points (0, 3) and (1, 8) on a graph paper and joining them, to form a line .

Q5 ) From the choices given below, choose the equation whose graphs are given in fig. (a) and fig. (b).

For fig. (a)	For fig. (b)
i) y = x	(i) y = x + 2
ii)x + y = 0	y = x - 2
iii)y = 2x	y = -x + 2
iv)2 + 3y = 7x	iv)x + 2y = 6

Answer :

In fig. (a), we can observe that the points(1, -1) and (-1, 1) lies on the line of equation x + y = 0.

Because, for (-1, 1), (1) + (-1) = 0

and for (1, -1), (-1) + (1) = 0


Similarly, in fig. (b), we can observe that the points(2, 0),(0, 2) and (-1, 3) lies on the line of equation x + y = 2.

Because, for (2, 0), (2) + (0) = 2,

for(0, 2), (0) + (2) = 2

And for (-1, 3), (-1) + (3) = 2

Q6 ) 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 units.

Answer :

Given that, work done by a body on application of a constant force is directly proportional to the distance travelled by the body.

i.e., (W) work done in distance (s) = F.s
Where,(F is an arbitrary constant which take the value 5 units)
Because, W = 5s.....(i)

s	W	Point
0	0	O(0, 0)
1	5	A(1, 5)
2	10	B(2, 10)

Now, plot the points on graph paper and join all the points to get a line.



i) From point B(2, 10), draw a line parallel to OY to intersect the x-axis at (2, 10) and draw a line parallel to OY to the x-axis intersect at C(0, 10).

work done by a body when distance travelled is 2 units = 10 units.


ii) Clearly y = 0, when x = 0, so the work done by a body when distance travelled is 0 units

Q7 ) 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 the contributions of Yamini and Fatima together towards the Prime Minster’s Relief Fund to help the earthquake victms are Rs. x and Rs. y, respectively.

Then, by given condition, x + y = 100 … (i)

x	0	100
y = 100 - x	100	0
(x, y)	B (0, 100)	A (100, 0)

Here, we plot the points B (0, 100) and A (100, 0) on graph paper and join all these points to form a line AB.

Q8 ) In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is linear equation that converts Fahrenheit to Celsius.
F = (\(\frac{9}{5} \) )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 \(30^{\circ}\)C , what is the temperature in Fahrenheit?
(iii) If the temperature is 95F , what is the temperature in Celsius?
(iv) If the temperature is \(0^{\circ}\)C , what is the temperature in Fahrenheit and if the temperature is 0F , 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 :

The linear equation that converts Fahrenheit to Celsius is
F = (\(\frac{9}{5} \))C + 32
\(\Rightarrow \) 5F - 9C = 160....(i)
For plotting points, we get the following table :

C	0	-160/9	-40
F = (9/5)C + 32	32	0	-40
Points	A(0, 32)	B(-160/9, 0)	c(-40, -40)

Here, we plot the given points on graph paper and join all these points to form a line.

ii)



ii)If the temperature is \(30^{\circ}\)C i.e.,C = \(30^{\circ}\)C
Then we get,

F = (\(\frac{9}{5} \) )×(30) + 32
= 9 × 6 + 32
= 54 + 32 = 86

Therefore, the temperature in Fahrenheit is 86F


iii)If the temperature is 95F i.e.,F = 95

Then we get, from eq.(i)

5(95) - 9C = 160
\(\Rightarrow \) 9C = 475 - 160 = 315

\(\therefore \) C = \(35^{\circ}\)C.

Therefore, the temperature in Celsius is \(35^{\circ}\)C


iv)If the temperature is \(0^{\circ}\)C i.e.,C = \(0^{\circ}\)

Then we get,

F = (\(\frac{9}{5} \) )×(0) + 32 = 32

Therefore, the temperature in Fahrenheit is 32F.

If the temperature is 0F i.e.,F = 0

Then we get, from eq.(i)

5(0) - 9C = 160
\(\Rightarrow \) C = \(\frac{-160}{9} ^{\circ}\)C

Therefore, the temperature in celsius is \(\frac{-160}{9} ^{\circ}\)C.


V)For this, let us take C = F,

Thus from eq.(i) we get,

5F - 9F = 160
\(\Rightarrow \) -4F = 160
\(\Rightarrow \) F = -40
Therefore, C = F = \(-40^{\circ}\)

Q1 ) Give the geometric representations of y = 3 as an equation
(i) In one variable
(ii) In two variables

Answer :

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)The representation of the solution on the number line is shown in the figure below, where y = 3 can be written as an equation in of two variable i.e., (0).x + y = 3

Here, if we observe, all the values of x are permissible as (0).x is always going to be 0.

However, y always has to be equal to 3.

Hence, graph of y = 3 will be a line parallel to x-axis and at a distance of 3 units along positive y- axis.

Q2 ) Give the geometric representations of 2x + 9 = 0 as an equation
(i) In one variable
(ii) In two variables

Answer :

The given linear equation is
\(\because \) 2x + 9 = 0
\(\Rightarrow x = \frac{-9}{2} \) ...(i)

(i) The representation of the solution on the number line is shown in the figure below, where x = \( \frac{-9}{2} \) is treated as an equation in one variable.




ii)The representation of the solution on the number line is shown in the figure below, where 2x + 9 = 0 can be written as an equation in of two variable i.e., 2x + (0).y = -9

Here, if we observe, all the values of y are permissible as (0).y is always going to be 0.

However, x always has to satisfy the equation to 2x + 9 = 0.

Hence, graph of x = \(\frac{-9}{2} \) will be a line parallel to y-axis and at a distance of \(\frac{-9}{2} \) units along negative x- axis.