NCERT Solutions for Class 9 Mathematics Chapter 4 – Linear equations in two variables

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

Answer

Let the cost of a notebook and a pen be x and y respectively.
Since, cost of a notebook = 2 x cost of a pen
x = 2y or x – 2y = 0
Which is required linear equation.

$\begin{array}{l}\text{Q.2 Express the following linear equations in the form}\\ \text{ax\hspace{0.17em}+\hspace{0.17em}by\hspace{0.17em}+\hspace{0.17em}c\hspace{0.17em}=\hspace{0.17em}0 and indicate the values of \hspace{0.17em}\hspace{0.17em}a,b\hspace{0.17em}\hspace{0.17em}and c in}\\ \text{each case:}\\ \left(\text{i}\right)2\mathrm{x}+3\mathrm{y}=9.3\overline{5}\left(\text{ii}\right)\mathrm{x}-\frac{\mathrm{y}}{5}-10=0\left(\text{iii}\right)-2\mathrm{x}+3\mathrm{y}=6\\ \left(\text{iv}\right)\mathrm{x}=3\mathrm{y}\left(\text{v}\right)2\mathrm{x}=-5\mathrm{y}\left(\text{vi}\right)3\mathrm{x}+2=0\\ \left(\text{vii}\right)\mathrm{y}-2=0\left(\text{viii}\right)5=2\mathrm{x}\end{array}$

Ans-

\begin{array}{l}\left(\text{i}\right)2x+3y=9.3\overline{5}\\ \text{or}2x+3y-9.3\overline{5}=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a = 2, b = 3 and c =}-9.3\overline{5}\\ \left(\text{ii}\right)x-\frac{y}{5}-10=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a = 1, b =}-\frac{1}{5}\text{and c =}-10\\ \left(\text{iii}\right)-2x+3y=6\\ \text{or}-2x+3y-6=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a =}-2\text{, b =}3\text{and c =}-6.\\ \left(\text{iv}\right)x=3y\\ \text{or}x-3y=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a =}1\text{, b =}-3\text{and c =}0.\\ \left(\text{v}\right)2x=-5y\\ \text{or}2x+5y=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=2\text{, b}=5\text{and c}=0.\\ \left(\text{vi}\right)3x+2=0\\ \text{or 3x}+0.y+2=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=3\text{, b}=0\text{and c}=2.\\ \left(\text{vii}\right)y-2=0\\ \text{or 0}\text{.x}+y-2=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=0\text{, b}=1\text{and c}=-2.\\ \left(\text{viii}\right)5=2x\\ \text{or}2x+0.y-5=0\\ \text{Comparing with ax + by + c = 0, we get}\\ \text{a}=2\text{, b}=0\text{and c}=-5.\end{array}

Q.3 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

Ans-

Since, y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. As for every value of x, there will be a value of y satisfying the above equation and vice-versa.

Therefore, the correct option is (iii).

\begin{array}{l}\text{Check which of the following are solutions of the equation}\\ x-\text{2y = 4 and which are not:}\\ \left(\text{i}\right)\left(\text{0,2}\right)\left(\text{ii}\right)\left(\text{2,0}\right)\left(\text{iii}\right)\left(\text{4,0}\right)\left(\text{iv}\right)\left(\sqrt{\text{2}}\text{,}\text{4}\sqrt{\text{2}}\right)\\ \left(\text{v}\right)\left(\text{1,1}\right)\end{array}

Q.4  MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C768@

Ans-

\begin{array}{l}\left(i\right)\left(0,2\right)\\ \text{Putting x}=0\text{and y}=2\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=0-2\left(2\right)=-4\ne \text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(\text{0,2}\right)\text{is not a solution of the given equation}\text{.}\\ \left(ii\right)\left(2,0\right)\\ \text{Putting x}=2\text{and y}=0\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=2-2\left(0\right)=2\ne \text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(2,0\right)\text{is not a solution of the given equation}\text{.}\\ \left(iii\right)\left(4,0\right)\\ \text{Putting x}=4\text{and y}=0\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=4-2\left(0\right)=4=\text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(\text{4,0}\right)\text{is a solution of the given equation}\text{.}\\ \left(iv\right)\left(\sqrt{2},4\sqrt{2}\right)\\ \text{Putting x}=\sqrt{2}\text{and y}=4\sqrt{2}\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y\\ =\sqrt{2}-2\left(4\sqrt{2}\right)\\ =-7\sqrt{2}\ne R.H.S\\ \text{Therefore,}\left(\sqrt{2},4\sqrt{2}\right)\text{is not a solution of the given equation}\text{.}\\ \left(v\right)\left(1,1\right)\\ \text{Putting x}=1\text{and y}=1\text{in L}\text{.H}\text{.S}\text{. of}x-2y=4,\text{}\text{we get}\\ \text{L}\text{.H}\text{.S}\text{.}=x-2y=1-2\left(1\right)=-1\ne \text{R}\text{.H}\text{.S}\\ \text{Therefore,}\left(\text{1,1}\right)\text{is not a solution of the given equation}\text{.}\end{array}

Q.5 Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Ans-

Putting x = 2 and y = 1 in 2x + 3y = k, we get
2(2) + 3(1) = k
4 + 3 = k
Or k = 7
Therefore the value of k is 7.

Q.6 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

Ans-

(i) Graph of x + y = 4 is given below:

(ii) Graph of x – y = 2 is given below:

(iii) Graph of y = 3x is given below:

(iv) Graph of 3 = 2x + y is given below:

Q.7 Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Ans-

The point (2, 14) satisfies the equations x + y = 16 and y – x = 12.
Therefore, x + y = 16 and y – x = 12 are the two lines passing through the point (2, 14).
Since, infinite number of lines can pass through one point, therefore, there are infinite lines of such type passing through the given point (2, 14).

Q.8 If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.

Ans-

Since, the point (3, 4) lies on the graph of the equation 3y = ax + 7, so it will satisfy the equation.
Putting x = 3 and y = 4 in given equation, we get
3(4) = a(3) + 7
12 = 3a + 7
5 = 3a
a = (5/3)

Q.9 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.

Ans-

Total covered distance = x km
Fare of first km distance = Rs. 8
Fare of the fare rest distance = (x – 1)5
Total fare of x km distance = 8 + (x – 1)5
y = 3 + 5x
or 5x – y + 3 = 0
which is the linear equation of the given informations.

The graph of 5x – y + 3 = 0 is given below:

Q.10 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 (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

Ans-

(i) Points on the given line are (−1, 1), (0, 0), and (1, −1).
We see that these points satisfies equation
x + y = 0. So, x + y =0 is the required equation of the given graph.
Hence, the correct answer is (ii) x + y = 0.

(ii) Since, points on the given line are (−1, 3), (0, 2), and (2, 0). We see that these points satisfies equation y = – x + 2. So, y = – x + 2 is the required equation of the given graph.
Therefore, the correct answer (iii) y = – x + 2.

Q.11 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.

Ans-

Let money donated by Yamini = Rs. x
And money donated by Fatima = Rs. y
Total contributed money = Rs. 100
Then, x + y = 100
The graph of linear equation x + y = 100 is:

Q.12 Give the geometric representations of y = 3 as an equation
(i) in one variable
(ii) in two variables

Ans-

(i) y = 3 in one variable can be represented as given below:

(ii) y = 3 in two variables can be represented as
0.x + y = 3
So, the solutions are:

The graphical representation of the line is as given below:

Q.13 Write four solutions for each of the following equations

(i) 2x + y = 7
(ii) πx + y = 9
(iii) x = 4y

Ans-

$\begin{array}{l}\left(\text{i}\right)\text{2x}+\text{y}=\text{7}\\ \mathrm{or}\mathrm{y}=7-2\mathrm{x}\\ \mathrm{We}\text{can get 4 solutions of the equation by putting x = 0, 1, 2, 3}\\ \text{y}=7-2\left(0\right)=7\\ \text{y}=7-2\left(1\right)=5\\ \text{y}=7-2\left(2\right)=3\\ \text{y}=7-2\left(3\right)=1\\ \mathrm{Therefore},\text{the four solutions of given equation are:}\\ (0,7),(1,5),(2,3)\text{and}(3,1).\\ \left(\text{ii}\right)\mathrm{\pi }\text{x}+\text{y}=\text{9}\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\text{x}\\ \mathrm{We}\text{can get 4 solutions of the equation by putting x = 0, 1, 2, 3}\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(0\right)=9\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(1\right)=9-\mathrm{\pi }\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(2\right)=9-2\mathrm{\pi }\\ \text{\hspace{0.17em}y}=9-\mathrm{\pi }\left(3\right)=9-3\mathrm{\pi }\\ \mathrm{Therefore},\text{the four solutions of given equation are:}\\ (0,9),(1,9-\mathrm{\pi }),(2,9-2\mathrm{\pi })\text{and}(3,9-3\mathrm{\pi }).\\ \left(\text{iii}\right)\text{x}=\text{4y}\\ \mathrm{We}\text{can get 4 solutions of the equation by putting y = 0, 1, 2, 3}\\ \text{x}=\text{4}\left(0\right)=0\\ \text{x}=\text{4}\left(1\right)=4\\ \text{x}=\text{4}\left(2\right)=8\\ \text{x}=\text{4}\left(3\right)=12\\ \mathrm{Therefore},\text{the four solutions of given equation are:}\\ (0,0),(1,4),(2,8)\text{and}(3,12).\end{array}$

Q.14 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

Ans-

$\begin{array}{l}\text{Let the distance travelled and the work done by the}\mathrm{}\text{body be x}\\ \text{and y respectively}.\\ \text{Work done}\mathrm{}\mathrm{\alpha }\text{\hspace{0.17em}\hspace{0.17em}distance travelled}\\ \Rightarrow \text{y}\mathrm{\alpha }\text{\hspace{0.17em}x}\\ \Rightarrow \text{y}=\text{kx}\mathrm{},\text{where k is constant.}\\ \mathrm{If}\text{contant force is 5 units, then work done}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}=5\mathrm{x}\\ \begin{array}{}\end{array}\end{array}$

(i) From the graphs, it can be observed that the value of y corresponding to x = 2 is 10. This implies that the work done by the body is 10 units when the distance travelled by it is 2 units.

(ii) From the graphs, it can be observed that the value of y corresponding to x = 0 is 0. This implies that the work done by the body is 0 units when the distance travelled by it is 0 unit.

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
Q.15 (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.

Ans-

(i) The given equation is: F = (9/5) C + 32
Let degree Celsius are taken on x-axis and Fahrenheit is taken on y- axis.
Then equation becomes:
y = (9/5) x + 32

The graph of the above equation is constructed as given below:

(ii) Temperature at 30°C
F = (9/5)30°C + 32
= 86
Therefore, temperature in Fahrenheit is 86°F.

(iii) Temperature = 95°F
95 = (9/5) C + 32
63 = (9/5) C
C = 63 × (5/9) = 35°
(iv) F = (9/5) C + 32
If C = 0°C, then
F = (9/5)(0) + 32
= 32
If F = 0, then
0 = (9/5) C + 32
–32 = (9/5) C
C = –32 x (5/9)
=–17.77
Therefore, if F = 0°F, then C = –17.77° C.
(v) F = (9/5) C + 32
If F = C, then
F = (9/5)F + 32
(1 – 9/5)F = 32
–(4/5)F = 32
F = 32 × (–5/4)
= – 40
Yes, there is a temperature, −40°, which is numerically the same in both Fahrenheit and Celsius.

Q.16 Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
(ii) in two variables

Ans-

(i) 2x + 9 = 0
⇒ x = – (9/2)
= – 4.5
2x + 9 = 0 in one variable can be represented as given below:

(ii) 2x + 9 = 0 in two variables can be represented as
2x + 0.y + 9 = 0
So, the solutions are:

The graphical representation of the line is as given below: