EXERCISE 4.1
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 ` x and that of a pen to be ` y).
Solution:
Let the cost of the notebook be x and the cost of a pen be y,
Then it is given that
Twice the cost of a pen = to the cost of a note book
2y=x
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
Solution:
2x+3y=9.35
subtracting 9.35 from both the LHS and the RHS to get
2x+3y-9.35 = 9.35 -9.35
=> 2x+3y - 9.35 =0
Now comparing it with the equation zx + by +c = 0 to get
a = 2
b = 3
and c = -9.35
(ii) x –5y– 10 = 0
Solution:
On comparing x –5y– 10 = 0 to ax+by+c=0 we get,
a=1, b= -5, and c=-10
(iii) –2x + 3y = 6
Solution:
–2x + 3y = 6 can be written as -2x+3y-6=0
And the required values are
a=-2, b= 3, and c=6
(iv) x = 3y
Solution:
x = 3y can be written as x-3y = 0
And the required values are
a = 1, b=-3 and c=0
(v) 2x = –5y
Solution:
2x = –5y can be written as 2x + 5y = 0
Thus, a = 2, b=5 and c=0
(vi) 3x + 2 = 0
Solution:
a = 3, b=0 and c=2
(vii) y – 2 = 0
Solution:
a=0, b=1 and c=-2
(viii) 5 = 2x
Solution:
5 = 2x can be written as 5-2x= 0 or 2x - 5 = 0
Thus
a=2 , b=0 and c=-5
EXERCISE 4.2
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
Answer:
(iii) infinitely many solutions
As we put different values of x we are sure to get a value of y.
For example: For x = -3,-2,-,1,0,1,2,3 the values of y are -4,-1,2,5,8,11 and so on.
2. Write four solutions for each of the following equations:
(i) 2x + y = 7
Solution: Four solution are:
(x=0,y=7)
(x=1,y=5)
(x=2,y=3)
(x=3,y=1)
(ii) πx + y = 9
Solution:
(x=0,y=9)
(x=1,y= 9 - π)
(x=2,y= 9-2π )
(x=3, y= 9-3π)
(iii) x = 4y
Solution:
(x=0,y=0)
(x=1,y=1/4)
(x=2,y=1/2)
(x=3,y=3/4)
3. Check which of the following are solutions of the equation x – 2y = 4 and which are
not:
(i) (0, 2)
Solution:
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=0 and y = 2, then
LHS = 0-4= - 4 ≠ RHS
Thus, Not a solution.
(ii) (2, 0)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=2 and y = 0, then
LHS = 2-0= 2 ≠ RHS
Thus, Not a solution.
(iii) (4, 0)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=4 and y = 0, then
LHS = 4-0= 4 = RHS
Thus, It is a solution.
(iv) (√2,4√2)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=√2, and y =4√2, then
LHS =√2 - 2(4√2) =√2 - 8√2) ≠ RHS
Thus, It is NOT a solution
.(v) (1, 1)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=1 and y = 1, then
LHS = 1-2(1)= -1 ≠ RHS
Thus, It is NOT a solution.
4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
It given that x=2 and y=1 is the solution of the equation 2x+3y=k
Therefore, if we put the values of x and y then the LHS and RHS of the above equation should be equal.
Therefore,
2(2) + 3(1) = k
=> 4 +3 =k
=> 7 = k
Thus k =7.
EXERCISE 4.3
1. Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
Solution:
First we will obtain two solution for this equation.
For x = 0 , y =4
For y = 0, x = 4
Now using these two solution (4,0) and (0,4) we graph can be plotted as
(ii) x – y = 2
Solution:
First we will obtain two solution for the above equation
put x= 0, then y =-2
put y=0 then x= 2
Thus the two solutions are (0,-2) and (2,0). Using these values the graph is
(iii) y = 3x
Solution:
First we will obtain two solution:
They are (0,0) and (1,3)
Using these two solution we can plot the line as
(iv) 3 = 2x + y
Solution:
The two solution for these equation are
(0,3) and (3/2,0). Using these we can plot the graph as
2. Give the equations of two lines passing through (2, 14). How many more such lines
are there, and why?
Solution:
Equations that can satisfy the above condition can be obtained from the equations 7x - y = K, Where k can be constant.
put k = 0, we get first equation as 7x - y= 0
put k = 1, we get second equation as 7x- y = 1 or 7x - y - 1 =0
And for that can pass through (2,14) are infinite.
3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
Given that the point (3,4) lies on the line 3y=ax+7. We put the values of x and y in there to get
3(4) = a(3) + 7
=> 12 = 3a +7
=> 3a = 5
=> a = 5/3
4. The taxi fare in a city is as follows: For the first kilometer, the fare is rupees 8 and for the
subsequent distance it is rupees 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.
Solution:
Given that the total fare is rupees y and the distance covered is x.
Now,
Rupees for the first kilometer = 8
And rate for the subsequent kilometers is (x-1)and amount is (x-1)5
And it is given that total amount is rs y and equation thus formed is
(x-1)5 + 8 = y
=> 5x- 5 + 8 = y
=> 5x +3 = y
=> 5x - y = 3.
To draw the graph, we first have to find atleast two solution point for the above equation
Put x=0 then y = -3
And put y= 0 then x = 3/5
Thus two points are (0,-3) and (3/5,0)
Therefore the graph is
5. 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
Fig. 4.6 Fig. 4.7
Solution:
For figure 4.6
We can observe that the line passes through point (0,0), (1,-,1) and (-1,1). Now we check equations for these points. The equations that has all these point is the correct one.
(i). x=y has point (0.0) but not (1,-1) and (-1,1). Therefore it is not the equation for fig 4.6
(ii) x+y = 0 has all the points lying on it, Thus is the equation for this figure. No need to check further.
For figure 4.7
We can observe that it passes through points (0,2),(2,0). And the only equations that has both these points is (iii) y = –x + 2. It is the required answer.
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:
Let the work done be x and the distance traveled be y under application of a constant force F.
It is Given that x proportional to y
=> x = Fy
Where, F is the constant force .
Thus , x= Fy is the required equation.
For the graph, given that the force is 5. Therefore the equation becomes x = 5y.
The points (0,0) and (5,1) lies on on this line.
(i) 2 units
Distance traveled is 2 units. Then Work done is = 5(2) = 10
(ii) 0 unit
Distance traveled is 0 units. Then worl Done is 5(0) = 0
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 ` x and
` y.) Draw the graph of the same.
Solution:
Let Yamini donated Rs. x
And Fatima donated Rs. y
Thus, their contribution is x + y = 100 And this is the required equation.
The graph for the same is
8. 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
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit
for y-axis.
Solution:
Let C = 0 then F = 32
C = -40 then F= -40
Therefore the graph is
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
Solution:
From the equation,
F = (9/5) C + 32
C= 30°C
Then F = (9/5) (30) + 32
= 54+32 = 86 Fahrenheit
(iii) If the temperature is 95°F, what is the temperature in Celsius?
Solution:
From equation
F = (9/5) C + 32 here F = 95°F
We get,
95 = (9/5)C + 32
=> (9/5) C = 63
=> C =35°C
(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?
Solution:
Using Equation F = (9/5) C + 32
Put C = 0°C Then F= 32°F
And Now Put F = 0°F Then
0 = (9/5)C + 32
=> (9/5)C = -32
=> C = (-32)(5/9)
=> C = -17.8 °C
(v) Is there a temperature which is numerically the same in both Fahrenheit and
Celsius? If yes, find it.
Solution :
In Equation F = (9/5) C + 32 put F=C
We have,
C = (9/5)C +32
=> (-4/5) C = 32
=> C= -40°C
Thus, At -40°C or -40 °F Both the temperatures are equal.
EXERCISE 4.4
1. Give the geometric representations of y = 3 as an equation
(i) in one variable
Solution: In one variable it represents a point on a number line.
In One variable y=3 represents a point on a number line.
(ii) in two variables
In two variable y = 3 means a line for whom the variable y is always 3.
Here it represent a line parallel to x axis and passing through the y axis at y=3.
2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
Solution:
It represents a point on a number line x= -4.5
(ii) in two variables
Solution:
It represents a line passing through x=--4.5 and parallel to y axis.
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 ` x and that of a pen to be ` y).
Solution:
Let the cost of the notebook be x and the cost of a pen be y,
Then it is given that
Twice the cost of a pen = to the cost of a note book
2y=x
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
Solution:
2x+3y=9.35
subtracting 9.35 from both the LHS and the RHS to get
2x+3y-9.35 = 9.35 -9.35
=> 2x+3y - 9.35 =0
Now comparing it with the equation zx + by +c = 0 to get
a = 2
b = 3
and c = -9.35
(ii) x –5y– 10 = 0
Solution:
On comparing x –5y– 10 = 0 to ax+by+c=0 we get,
a=1, b= -5, and c=-10
(iii) –2x + 3y = 6
Solution:
–2x + 3y = 6 can be written as -2x+3y-6=0
And the required values are
a=-2, b= 3, and c=6
(iv) x = 3y
Solution:
x = 3y can be written as x-3y = 0
And the required values are
a = 1, b=-3 and c=0
(v) 2x = –5y
Solution:
2x = –5y can be written as 2x + 5y = 0
Thus, a = 2, b=5 and c=0
(vi) 3x + 2 = 0
Solution:
a = 3, b=0 and c=2
(vii) y – 2 = 0
Solution:
a=0, b=1 and c=-2
(viii) 5 = 2x
Solution:
5 = 2x can be written as 5-2x= 0 or 2x - 5 = 0
Thus
a=2 , b=0 and c=-5
EXERCISE 4.2
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
Answer:
(iii) infinitely many solutions
As we put different values of x we are sure to get a value of y.
For example: For x = -3,-2,-,1,0,1,2,3 the values of y are -4,-1,2,5,8,11 and so on.
2. Write four solutions for each of the following equations:
(i) 2x + y = 7
Solution: Four solution are:
(x=0,y=7)
(x=1,y=5)
(x=2,y=3)
(x=3,y=1)
(ii) πx + y = 9
Solution:
(x=0,y=9)
(x=1,y= 9 - π)
(x=2,y= 9-2π )
(x=3, y= 9-3π)
(iii) x = 4y
Solution:
(x=0,y=0)
(x=1,y=1/4)
(x=2,y=1/2)
(x=3,y=3/4)
3. Check which of the following are solutions of the equation x – 2y = 4 and which are
not:
(i) (0, 2)
Solution:
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=0 and y = 2, then
LHS = 0-4= - 4 ≠ RHS
Thus, Not a solution.
(ii) (2, 0)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=2 and y = 0, then
LHS = 2-0= 2 ≠ RHS
Thus, Not a solution.
(iii) (4, 0)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=4 and y = 0, then
LHS = 4-0= 4 = RHS
Thus, It is a solution.
(iv) (√2,4√2)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=√2, and y =4√2, then
LHS =√2 - 2(4√2) =√2 - 8√2) ≠ RHS
Thus, It is NOT a solution
.(v) (1, 1)
Given equation x – 2y = 4
LHS = x-2y
RHS = 4
Put x=1 and y = 1, then
LHS = 1-2(1)= -1 ≠ RHS
Thus, It is NOT a solution.
4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Solution:
It given that x=2 and y=1 is the solution of the equation 2x+3y=k
Therefore, if we put the values of x and y then the LHS and RHS of the above equation should be equal.
Therefore,
2(2) + 3(1) = k
=> 4 +3 =k
=> 7 = k
Thus k =7.
EXERCISE 4.3
1. Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
Solution:
First we will obtain two solution for this equation.
For x = 0 , y =4
For y = 0, x = 4
Now using these two solution (4,0) and (0,4) we graph can be plotted as
(ii) x – y = 2
Solution:
First we will obtain two solution for the above equation
put x= 0, then y =-2
put y=0 then x= 2
Thus the two solutions are (0,-2) and (2,0). Using these values the graph is
(iii) y = 3x
Solution:
First we will obtain two solution:
They are (0,0) and (1,3)
Using these two solution we can plot the line as
(iv) 3 = 2x + y
Solution:
The two solution for these equation are
(0,3) and (3/2,0). Using these we can plot the graph as
2. Give the equations of two lines passing through (2, 14). How many more such lines
are there, and why?
Solution:
Equations that can satisfy the above condition can be obtained from the equations 7x - y = K, Where k can be constant.
put k = 0, we get first equation as 7x - y= 0
put k = 1, we get second equation as 7x- y = 1 or 7x - y - 1 =0
And for that can pass through (2,14) are infinite.
3. If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Solution:
Given that the point (3,4) lies on the line 3y=ax+7. We put the values of x and y in there to get
3(4) = a(3) + 7
=> 12 = 3a +7
=> 3a = 5
=> a = 5/3
4. The taxi fare in a city is as follows: For the first kilometer, the fare is rupees 8 and for the
subsequent distance it is rupees 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.
Solution:
Given that the total fare is rupees y and the distance covered is x.
Now,
Rupees for the first kilometer = 8
And rate for the subsequent kilometers is (x-1)and amount is (x-1)5
And it is given that total amount is rs y and equation thus formed is
(x-1)5 + 8 = y
=> 5x- 5 + 8 = y
=> 5x +3 = y
=> 5x - y = 3.
To draw the graph, we first have to find atleast two solution point for the above equation
Put x=0 then y = -3
And put y= 0 then x = 3/5
Thus two points are (0,-3) and (3/5,0)
Therefore the graph is
5. 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
Fig. 4.6 Fig. 4.7
Solution:
For figure 4.6
We can observe that the line passes through point (0,0), (1,-,1) and (-1,1). Now we check equations for these points. The equations that has all these point is the correct one.
(i). x=y has point (0.0) but not (1,-1) and (-1,1). Therefore it is not the equation for fig 4.6
(ii) x+y = 0 has all the points lying on it, Thus is the equation for this figure. No need to check further.
For figure 4.7
We can observe that it passes through points (0,2),(2,0). And the only equations that has both these points is (iii) y = –x + 2. It is the required answer.
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:
Let the work done be x and the distance traveled be y under application of a constant force F.
It is Given that x proportional to y
=> x = Fy
Where, F is the constant force .
Thus , x= Fy is the required equation.
For the graph, given that the force is 5. Therefore the equation becomes x = 5y.
The points (0,0) and (5,1) lies on on this line.
(i) 2 units
Distance traveled is 2 units. Then Work done is = 5(2) = 10
(ii) 0 unit
Distance traveled is 0 units. Then worl Done is 5(0) = 0
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 ` x and
` y.) Draw the graph of the same.
Solution:
Let Yamini donated Rs. x
And Fatima donated Rs. y
Thus, their contribution is x + y = 100 And this is the required equation.
The graph for the same is
8. 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
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit
for y-axis.
Solution:
Let C = 0 then F = 32
C = -40 then F= -40
Therefore the graph is
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
Solution:
From the equation,
F = (9/5) C + 32
C= 30°C
Then F = (9/5) (30) + 32
= 54+32 = 86 Fahrenheit
(iii) If the temperature is 95°F, what is the temperature in Celsius?
Solution:
From equation
F = (9/5) C + 32 here F = 95°F
We get,
95 = (9/5)C + 32
=> (9/5) C = 63
=> C =35°C
(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?
Solution:
Using Equation F = (9/5) C + 32
Put C = 0°C Then F= 32°F
And Now Put F = 0°F Then
0 = (9/5)C + 32
=> (9/5)C = -32
=> C = (-32)(5/9)
=> C = -17.8 °C
(v) Is there a temperature which is numerically the same in both Fahrenheit and
Celsius? If yes, find it.
Solution :
In Equation F = (9/5) C + 32 put F=C
We have,
C = (9/5)C +32
=> (-4/5) C = 32
=> C= -40°C
Thus, At -40°C or -40 °F Both the temperatures are equal.
EXERCISE 4.4
1. Give the geometric representations of y = 3 as an equation
(i) in one variable
Solution: In one variable it represents a point on a number line.
(ii) in two variables
In two variable y = 3 means a line for whom the variable y is always 3.
Here it represent a line parallel to x axis and passing through the y axis at y=3.
2. Give the geometric representations of 2x + 9 = 0 as an equation
(i) in one variable
Solution:
It represents a point on a number line x= -4.5
(ii) in two variables
Solution:
It represents a line passing through x=--4.5 and parallel to y axis.
No comments:
Post a Comment