NCERT Solutions for Exercise 4.3 Class 9 Maths Chapter 4 - Linear Equations in Two Variables

Q1 (ii) Draw the graph of each of the following linear equations in two variables: $x - y = 2$

Answer:

Given : $x - y = 2$

Putting x=0,we have $y=0-2=-2$

Putting x=1,we have $y=1-2=-1$

Thus, (0,-2) and (1,-1) are solutions of given equation.

Q1 (iii) Draw the graph of each of the following linear equations in two variables: $y = 3x$

Answer:

Given : $y = 3x$

Putting x=0,we have $y=3(0)=0$

Putting x=1,we have $y=3(1)=3$

Thus, (0,0) and (1,3) are solutions of given equation.

Q1 (iv) Draw the graph of each of the following linear equations in two variables: $3 = 2x + y$

Answer:

Given : $3 = 2x + y$

Putting x=0,we have $y=3-2.(0)=3$

Putting x=1,we have $y=3-2.(1)=1$

Thus, (0,3) and (1,1) are solutions of given equation.

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

Answer:

The equations of two lines passing through (2, 14) are given by : $x+y=16\, \, and\, \, x-y=-12$

There are infinite lines passing through (2, 14) because infinite lines pass through a point.

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

Answer:

Given : the point (3, 4) lies on the graph of the equation $3y = ax+ 7$

Put x=3 and y=4

$3y = ax+ 7$

$\Rightarrow 3(4) = a(3)+ 7$

$\Rightarrow 12 = 3a+ 7$

$\Rightarrow 12 -7 = 3a$

$\Rightarrow 5 = 3a$

$\Rightarrow a=\frac{5}{3}$

Q4 The taxi fare in a city is as follows: For the first 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.

Answer:

Given: The distance covered as x km and total fare is Rs. y.

Total fare =the fare for first km + the fare of the remaining distance

$\therefore y=8+5\times (x-1)$

$\Rightarrow y=8+5x-5$

$\Rightarrow y=3+5x$

For graph,

Putting x=0, we have $y=3+5(0)=3$

Putting x=1, we have $y=3+5(1)=8$

Putting x=2, we have $y=3+5(2)=13$

Hence,(0,3),(1,8) and (2,13) are solutions of equation.

The graph is as shown :

Q5 (A) From the choices given below, choose the equation whose graph is given in Fig. 4.6.

(i) $y = x$

(ii) $x + y = 0$

(iii) $y = 2x$

(iv) $2 + 3 y = 7x$

Answer:

For the given figure :

Points on line are (-1,1) , (0,0 ) and (1,-1)

$x + y = 0$ satisfies all the above points.

Thus, $x + y = 0$ is the correct equation of the line.

Q5 (B) From the choices given below, choose the equation whose graph is given in Fig. 4.7

(i) $y = x + 2$

(ii) $y = x - 2$

(iii) $y = -x + 2$

(iv) $x + 2y = 6$

Answer:

For the given figure :

Points on line are (-1,3) , (0,2 ) and (2,0)

$y = -x + 2$ satisfies all the above points.

Thus, $y = -x + 2$ is the correct equation of the line.

Q6 (i) 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: 2 units

Answer:

Let work done be y and distance be x.

Given : Constant force = 5 units

Work done by a body on the application of a constant force is directly proportional to the distance travelled by the body.

i.e. $y\, \, \alpha \, \, x$

$\Rightarrow y=kx$

k=Constant force = 5

Then,

$\Rightarrow y=5x$

For graph,

Put x=0,we have $y=5(0)=0$

Put x=1,we have $y=5(1)=5$

Put x=2,we have $y=5(2)=10$

Points are (0,0) , (0,5) and (2,10)

If the distance travelled is 2 units then the work done is 10 units.

Q6 (ii) 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 : 0 unit

Answer:

Let work done be y and distance be x.

Given : Constant force = 5 units

Work done by a body on the application of a constant force is directly proportional to the distance travelled by the body.

i.e. $y\, \, \alpha \, \, x$

$\Rightarrow y=kx$

k=Constant force = 5

Then, $\Rightarrow y=5x$

For graph,

Put x=0,we have $y=5(0)=0$

Put x=1,we have $y=5(1)=5$

Put x=2,we have $y=5(2)=10$

Points are (0,0) , (0,5) and (2,10)

If the distance travelled is 0 units then work done 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 contribution of Yamini be x.

contribution of Yamini be y.

According to question,

$x+y=100$

For x=0 , we have $y=100-0=100$

For x=10 , we have $y=100-10=90$

For x=20 , we have $y=100-20=80$

Hence, (0,100) , (10,90) and (20,80)

Q8 (i) In countries like the USA and Canada, the 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: $F = \left(\frac{9}{5} \right )C + 32$

Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.

Answer:

Let celsius be on x-axis and Fahrenheit be on y-axis.

$F = \left(\frac{9}{5} \right )C + 32$

For graph,

Putting x=0, we get $y= \left(\frac{9}{5} \right )(0)+ 32=32$

Putting x=5, we get $y= \left(\frac{9}{5} \right )(5)+ 32=41$

Putting x=10, we get $y= \left(\frac{9}{5} \right )(10)+ 32=50$

Hence, points are (0,32) , (5,41) and (10,50).

Q8 (ii) 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: $F = \left(\frac{9}{5} \right )C + 32$

If the temperature is 30°C, what is the temperature in Fahrenheit?

Answer:

$F = \left(\frac{9}{5} \right )C + 32$

Put c=30,

$F = \left(\frac{9}{5} \right )30 + 32=54+32=86$

Thus, the temperature = 30°C, then the temperature is 86 in Fahrenheit.

Q8 (iii) In countries like USA and Canada, the 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: $F = \left(\frac{9}{5} \right )C + 32$

If the temperature is 95°F, what is the temperature in Celsius?

Answer:

$F = \left(\frac{9}{5} \right )C + 32$

Put F=95,

$95 = \left(\frac{9}{5} \right )C + 32$

$\Rightarrow \left(\frac{9}{5} \right )C=95 -32$

$\Rightarrow \left(\frac{9}{5} \right )C=63$

$\Rightarrow C=35$

If the temperature is 95°F, then 35 is the temperature in Celsius.

Q8 (iv) 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: $F = \left(\frac{9}{5} \right )C + 32$

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?

Answer:

If the temperature is 0°C,

$F = \left(\frac{9}{5} \right )C + 32$

$F = \left(\frac{9}{5} \right )(0) + 32$

$\Rightarrow F = 32$

if the temperature is 0°F,

$F = \left(\frac{9}{5} \right )C + 32$

$0= \left(\frac{9}{5} \right )C + 32$

$\Rightarrow \left(\frac{9}{5} \right )C=- 32$

$\Rightarrow C=- 17.8$

Thus, if the temperature is 0°C , then the temperature in Fahrenheit is 32 and if the temperature is 0°F, then the temperature in celsius is -17.8.

Q8 (v) 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: $F = \left(\frac{9}{5} \right )C + 32$

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Answer:

$F = \left(\frac{9}{5} \right )C + 32$

Let temperature be x in both Fahrenheit and celsius.

$x= \left(\frac{9}{5} \right )x + 32$

$\Rightarrow 5x= 9x + 160$

$\Rightarrow 5x-9x = 160$

$\Rightarrow -4x = 160$

$\Rightarrow x = -40 ^\circ$

Thus, 40 is the temperature which is numerically the same in both Fahrenheit and celsius.