Question-1
Solution:
Let the cost of a note book be Rs x and the cost of a pen be Rs y.
Then, according to the given condition of the question,
x = 2y
â‡’ x â€“ 2y = 0
This is the required linear equation in two variables x and y.
Question-2
(i) 2x + 3y = 9.35 (ii) x - - 10 = 0
Solution:
(i) 2x + 3y = 9.35
â‡’ 2x + 3y - 9.35 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = 3, c = -9.35.
(ii) x - -10 = 0
â‡’ 5x â€“ y - 50 = 0 (Multiplying both sides by 5)
Comparing with ax + by + c = 0, we get
a = 5, b = -1, c = -50
Question-3
(i) â€“ 2x + 3y = 6 (ii) x = 3y.
Solution:
(i) â€“ 2x + 3y = 6
â€“ 2x + 3y â€“ 6 = 0
Comparing with ax + by + c = 0, we get
a = â€“ 2, b = 3, c = -6.
Another Method:
â€“ 2x + 3y = 6
â‡’ â€“ 2x + 3y â€“ 6 = 0
â‡’ 2x â€“ 3y + 6 = 0
Comparing with ax + by + c = 0, we get (Multiplying both sides by â€“ 1)
a = 2, b = â€“ 3, c = 6.
(ii) x = 3y
â‡’ x â€“ 3y + 0 = 0
Comparing with ax + by + c = 0, we get
a = 1, b = â€“ 3, c = 0
Question-4
(i) 2x = â€“ 5y (ii) 3x + 2 = 0
Solution:
(i) 2x = â€“ 5y
2x = â€“ 5y
â‡’ 2x + 5y + 0 = 0
Comparing with ax + by + c = 0, we get
a = 2, b = 5, c = 0.
(ii) 3x + 2 = 0
3x + 2 = 0
â‡’ 3x + 0y + 2 = 0
Comparing with ax + by + c = 0, we get
a = 3, b = 0, c = 2.
Question-5
(i) y â€“ 2 = 0 (ii) 5 = 2x.
Solution:
(i) y â€“ 2 = 0
Comparing with ax+by+c = 0
(0)x + (1)y + ( â€“ 2) = 0
Therefore
a = 0, b = 1, c = â€“ 2
(ii) 5 = 2x
Comparing with ax + by + c = 0
2x â€“ (0)y â€“ 5 = 0, we get
a = 2, b = 0, c = â€“ 5.
Question-6
(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions
Solution:
The option (iii) y = 3x + 5 has infinitely many solution is true.
Reason.
For every value of x, there is a corresponding value of y and vice-versa.
Question-7
(i) 2x + y = 7 (ii) Ï€x + y = 9
Solution:
(i) 2x + y = 7
2x + y = 7
â‡’y = 7 â€“ 2x
Put x = 0, we get y = 7 â€“ 2(0) = 7 â€“ 0 = 7
Put x = 1, we get y = 7 â€“ 2(1) = 7 â€“ 2 = 5
Put x = 2, we get y = 7 â€“ 2(2) = 7 â€“ 4 = 3
Put x = 3, we get y = 7 â€“ 2(3) = 7 â€“ 6 = 1
âˆ´ Four solutions are (0,7), (1, 5), (2, 3) and (3, 1).
(ii) Ï€ x + y = 9
y = 9 - Ï€ x
Put x = 0, we get y = 9 â€“ Ï€ (0) = 9 â€“ 0 = 9
Put x = 1, we get y = 9 â€“ Ï€ (1) = 9 â€“ Ï€
Put x = â€“ 1, we get y = 9 â€“ Ï€ (-1) = 9 +Ï€
Put x = , we get y = 9 â€“ Ï€ = 9 â€“ 9 = 0
âˆ´ Four solutions are (0,9), (1, 9 -Ï€ ), ( â€“ 1, 9 + Ï€) and
Question-8
Solution:
x = 4y
â‡’ y =
Put x = 0, we get y == 0
Put x = 4, we get y == 1
Put x = - 4, we get y = = -1
Put x = 2, we get y ==
Four solutions are (0, 0), (4, 1), (- 4, - 1) and
Question-9
(i) (0, 2) (ii) (2, 0)
Solution:
The given equation is x â€“ 2y = 4 ...(1)
(i) (0, 2)
Put x = 0 and y = 2 in (1), we get
x - 2y = 0 - 2(2) = â€“ 4, which is not 4.
âˆ´ (0, 2) is not a solution of (1).
(ii) (2, 0)
Put x = 2 and y = 0 in (1), we get
x â€“ 2y = 2 â€“ 2(0) = 2 â€“ 0 = 2, which is not 4.
âˆ´ (2, 0) is not a solution of (1).
Question-10
(i) (4, 0)
(ii) (, 4) (iii) (1, 1).
Solution:
x â€“ 2y = 4 ------ (1)
(i) (4, 0)
Put x = 4 and y = 0 in (1), we get
x â€“ 2y = 4 â€“ 2(0) = 4
âˆ´ (4, 0) is a solution of (1).
(ii) (, 4)
Put x =, y = 4 in (1), we get
x â€“ 2y = â€“ 2(4) â‡’ â€“ 8 = â€“ 7 which is not 4.
âˆ´ (, 4 ) is not a solution of (1).
(iii) (1,1)
Put x = 1 and y = 1 in (1), we get
x - 2y = 1 â€“ 2(1) = 1 â€“ 2 = â€“ 1, which is not 4.
âˆ´ (1, 1) is not a solution of (1).
Question-11
Solution:
If x = 2, y = 1 is a solution of the equation 2x + 3y = k, then these values will satisfy the equation So, putting x = 2 and y = 1 in the equation, we get
2(2) + 3(1) = k
â‡’ 4 + 3 = k
â‡’ k = 7
Question-12
(i) x + y = 4 (ii) x â€“ y = 2
Solution:
(i) x + y = 4
â‡’ y = 4 - x
Table of solutions
x |
1 |
2 |
y |
3 |
2 |
We plot the points(1, 3) and (2, 2) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x + y = 4.
(ii) x â€“ y = 2
x â€“ y = 2
â‡’ y = x - 2
Table of solutions
x | 2 | 3 |
y | 0 | 1 |
We plot the points(2, 0) and (3, 1) on the graph paper and join the same by a ruler to get the line which is the graph of the equation x â€“ y = 2.
Question-13
Solution:
y = 3x
Table of solutions
x | 0 | 1 |
y | 0 | 3 |
We plot the point (0, 0) and (1, 3) on the graph paper and join the same by a ruler to get the line which is the graph of the equation y = 3x .
Question-14
Solution:
3 = 2x + y
â‡’ y = 3 â€“ 2 x
Table of solutions
x | 1 | 0 |
y | 1 | 3 |
We plot the points (1, 1) and (0, 3) on the graph paper and join the same by an ruler to get the line which is the graph of the equation 3 = 2 x + y.
Question-15
lines are there, and why?
Solution:
Question-16
of a.
Solution:
3y = ax + 7, then
3(4) = a(3) + 7
â‡’ 12 = 3a + 7
â‡’ 3a = 12 â€“ 7
â‡’ 3a = 5
â‡’ a =
Question-17
for the subsequent distance it is Rs 5 per km. Taking the distance covered as
x km and total far as Rs y, Write a linear equation for this information, and
draw its graph.
Solution:
Total distance covered = x km
Total fare = Rs y
Fair for the first kilometer = Rs.8
Subsequent distance = (x â€“ 1) km
Fair for the subsequent distance = Rs 5(x â€“ 1)
According to the question,
y = 8 + 5(x â€“ 1)
â‡’ y = 8 + 5x â€“ 5
â‡’ y = 5x + 3
Table of solutions
x | 0 | 1 |
y | 3 | 8 |
We plot the points(0, 3) and (1, 8) on the graph paper and join the same by a ruler to get the line which is the graph of the equation
y = 5x + 3
Question-18
From the choices given below, choose the equation whose graphs are given in
Fig. (1) and Fig. (2).
(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
Solution:
For Fig. (1). The Correct equation is (ii) x + y = 0
For Fig. (2). The Correct equation is (iii) y = -x + 2
Question-19
(i) 2 units (ii) 0 units
Solution:
Let the work done by the constant force by y units and the distance traveled by the body be x units.
Constant force = 5 units
We know that
Work done = Force Â´ Displacement
â‡’y = 5x
Table of solutions
x | 0 | 1 |
y | 0 | 5 |
We plot the points(0, 0) and (1, 5) on the graph paper and join the same by a ruler to get the line which is the graph of equation y = 5x.
Work done when the distance travelled is 2 units = 10 units
Work done when the distance travelled is 0 units = 0 units
Question-20
Rs.100 towards the Prime Ministerâ€™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 be Rs x and Rs y respectively.
Then according to the question
x + y = 100
This is the linear equation which the given data satisfies
Now, x + y = 100
â‡’y = 100 â€“ x
Table of solutions
x | 0 | 50 |
y | 100 | 50 |
We plot the points(0, 100) and (50, 50) on the graph paper and join the same by a ruler to get the line which is the graph of equation
x + y = 100.
Question-21
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 â€“ axis and
Fahrenheit for y-axis.
(ii) If the temperature is 30^{0 }C, What is the temperature in Fahrenheit?
Solution:
(i) F =
We draw the X and Y axis with the each X co-ordinate = 10 units and each Y co-ordinate = 10 units
(ii) F =
When C = 30, then F =
= 54 + 32
= 86
Question-22
Whereas in countries like India. It is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
F = C + 32
(i) If the temperature is 95Â° ^{ }F, What is the temperature in Celsius?
(ii) If the temperature is 0Â° C, What is the temperature in Fahrenheit and if the temperature is 0Â° F, What is the temperature in Celsius?
Solution:
(i) When F = 95, then
F =
â‡’95 =
â‡’
â‡’
â‡’ C =
â‡’ C = 35
âˆ´ Required temperature = 35Â°^{ }C
(ii) When C = 0, then F =
âˆ´ Required temperature = 32Â° ^{ }F
Whan F = 0, then
F =
â‡’0 =
â‡’
â‡’ C =
= -
âˆ´ Required temperature = -^{Â°}^{ }C
Question-23
Whereas in countries like India. It is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
F = C + 32
Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it?
Solution:
Let the temperature be x numerically. Then,
â‡’ x = -
âˆ´ Numerical value of required temperature = -40
Question-24
Solution:
The given equation is y = 3
In one variable
The representation of y = 3 on the number line is as shown below:
Question-25
Solution:
In two variables
y = 3
â‡’ 0.x + 1.y = 3
It is a linear equation in two variables x and y. This is represented by a line. All the values of x are permissible because 0Ã—x is always 0. However, y must satisfy the relation y = 3. Hence, two solutions of the given equation are x = 0, x = 2, y = 3
Thus the graph AB is a line parallel to the x-axis at a distance of 5 units above it.
Question-26
Solution:
The given equation is 2x + 9 = 0
In one variable
2x + 9 = 0 â‡’ 2x = -9 â‡’ x = -
The representation of 2x +9 on the number line is as shown below:
Question-27
Solution:
In two variables
2x + 9 = 0 â‡’ 2x + 0y + 9 = 0
It is a linear equation in two variables x and y. This is represented by a line. All the values of y are permissible because 0y is always 0. However, x must satisfy the relation 2x + 9 = 0, i.e. x = -, y = 2
The graph AB is a line parallel to the y-axis and at a distance units to the left of it.