Let the present age of Aftab and his father be x years and y years respectively.
According to question,
7 years ago, we have
x – 7 = 7 (y - 7)
Or, x – 7 = 7y – 49
Or, x – 7y = - 42 …………… (1)
3 years from now, we have
(x + 3) = 3 (y + 3)
Or, x + 3 = 3y + 9
Or, x – 3y = 6 ……………. (2)
Graphical Representation
From equation (1), x – 7y = -42
Table value of x and y
|
x: |
0 |
-42 |
-35 |
|
y: |
6 |
0 |
1 |
From equation (2), x – 3y = 6
Table value of x and y
|
x: |
0 |
9 |
6 |
|
y: |
-2 |
1 |
0 |
Plotting the tables on the graph:
Q:
Let the cost of one bat and one ball be x Rupees and y Rupees respectively.
According to first condition,
3x + 6y = ₨ 3900 …………….(1)
According to second condition,
x + 3y = ₨ 1300 …………….(2)
Graphical Representation
From equation (1), 3x + 6y = 3900
Table value of x and y
|
x: |
100 |
300 |
700 |
|
y: |
600 |
500 |
300 |
From equation (2), x + 3y = 1300
Table value of x and y
|
x: |
100 |
700 |
400 |
|
y: |
400 |
200 |
300 |
1 Unit = 100
Q:
Let the cost of one kg apple be x ₨ and 1 kg grapes be y ₨.
According to first condition,
2x + y = 160 ₨ …………..(1)
According to second condition,
4x + 2y = 300
2x + y = 150 ……………….(2)
Graphical Representation
Table for equation (1), 2x + y = 160
Table value of x and y
|
x: |
40 |
60 |
80 |
|
y: |
60 |
40 |
0 |
Table for equation (2), 2x + y = 150
Table value of x and y
|
x: |
40 |
60 |
20 |
|
y: |
70 |
30 |
110 |
Exercise 3
(i) x + y = 14 …………….(1)
x – y = 4 …………….(2)
From the equation (1), we get
x = 14 - y …………….(3)
Putting the value of x in equation (2), we get
(14 - y) – y = 4
14 – y – y = 4
- 2y = - 10
Putting the value of y in equation (3),
x = 14 – 5
x = 9
Hence, x = 9 and y = 5
(iii) 3x - y = 3 …………….(1)
9x – 3y = 9…………….(2)
From the equation (1), we get
Putting the value of y in equation (2), we get
9x – 3 (3x - 3) = 9
9x – 9x + 9 = 9
9 = 9, which is true.
Therefore, pair of linear equation has infinite many solutions.
Q:
(i) Let the numbers be x and y, such that x > y
Therefore, according to question
x - y = 26 …………….(1)
x = 3y…………….(2)
Putting the value of x from equation (2) to equation (1), we get
3y – y = 26
2y = 26
y = 13
Putting the value in equation (2), we get
x = 3 x 13
x = 39
Hence, the numbers are 39 and 13.
(ii) Let one be x◦ and other be y◦ such that (x◦ > y◦)
Therefore, according to question
x◦ + y◦ = 180◦…………….(1) (Supplementary angles)
x◦ = 18 + y◦…………….(2)
Putting the value of x from equation (2) to equation (1), we get
18 + y◦ + y◦ = 180◦
2y◦ = 162◦
Putting the value of y in equation (2), we get
x◦ = 18 + 81
x = 99
(iv) Let the fixed charge be = ₨ x
Let the charge for 1 km distance be = ₨ y
According to first condition,
x + 10y = ₨ 105
x = 105 – 10y …………….(1)
According to second condition,
x + 15y = 155 ………………(2)
Putting the value of x in equation (2), we get
105 – 10y + 15y = 155
5y = 50
y = 10
Putting the value of y in equation (2), we get
x + 15 x 10 = 155
x = 5
Hence, the fixed charge for taxi is ₨ 5 and, the charge for one km distance is ₨ 50.
Charge for 25 km distance
= 25 x 10 + 5
= ₨ 255
(vi) Let the age of Jacob be = x years
Let the age of Jacob’s father be = y years
After 5 years,
Jacob’s age x + 5 years
Son’s age y + 5 years
According to question,
x + 5 = 3 (y + 5)
x + 5 = 3y + 15
x = 3y + 10………………(1)
Five years ago,
(x- 5) = 7 (y - 5)
x – 5 = 7y – 35
x – 7y = -30……………….(2)
Putting the value of x in equation (2), we get
3y + 10 - 7y = -30
-4y = -40
y = 10
Putting the value of y in equation (1), we get
x = 3 (10) + 10
x = 40
Hence, the present age of Jacob is 40 years and the age of his son is 10 years.
Given equations,
x + y + 1 = 0 …………………. (1)
3x + 2y + 2 = 0 ………………….. (2)
From the equation (1), we get
|
x |
0 |
1 |
2 |
|
y |
1 |
2 |
3 |
From the equation (2), we get
|
x |
4 |
3 |
0 |
|
y |
0 |
3 |
6 |
The coordinates of vertices of triangle formed by these lines and the x- axis are (-1, 0), (4, 0), (2, 3).
, , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 
Find the fraction.
