Linear Equations
NCERT Exercise 3.1
Question 1: Aftab tells his daughter, Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.
Represent this situation algebraically and graphically.
Answer: Let us assume Aftab’s current age = x and his daughter’s current age = y
Seven years ago:
Aftab’s age = x - 7 and daughter’s age = y - 7
As per question, seven years ago, Aftab was seven times as old as his daughter was seven years ago.
So, x – 7 = 7(y – 7)
Solving this equation, we get
x – 7 = 7y – 49
Or, x = 7y – 49 + 7
Or, x = 7y – 42
Or, 7y – x – 42 = 0 ……(1)
This equation gives following values for x and y
x = - 35, - 28, - 21, - 14, - 7, 0, 7, 14, 21, 28, 35, 42, 49
y = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
Three years from now:
Aftab’s age = x + 3 and daughter’s age = y + 3
As per question, three years from now, Aftab will be thtice as old as his daughter.
So, x + 3 = 3(y + 3)
Or, x + 3 = 3y + 9
Or, x = 3y + 9 – 3
Or, x = 3y + 6
Or, 3y – x + 6 = 0 ……..(2)
This equation gives following values for x and y:
x = 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45
y = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
The following graph is plotted for the given pair of linear equations:
The lines for x and y intersect at point representing the coordinates (12, 42)
So, daughter’s age = 12 years and Aftab’s age = 42 years
Question 2: The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Answer: Let us assume that price of one bat = x and price of one ball = y. Following equations can be written as per the question:
Price of 3 bats and 6 balls:
3x + 6y = 3900
Or, x + 2y = 1300 ……..(1)
This equation will give following values for x and y
x | 1500, 1400, 1300, 1200, 1100 |
y | - 100, - 50, 0, 50, 100 |
Now, price of 1 bat and 3 balls:
x + 3y = 1300 ……….(2)
This equation will give following values for x and y
x | 1600, 1450, 1300, 1150, 1000 |
y | - 100, - 50, 0, 50, 100 |
The following graph is plotted for the given pair of linear equations.
The lines for x and y intersect at point representing the coordinates (1300, 0)
So, Price of one bat = Rs. 1300 and Price of one ball = zero
Question 3: The cost of 2 kg apples and 1 kg grapes was found to be Rs. 160. After a month, the cost of 4 kg apples and 2 kg grapes is Rs. 300. Represent this situation algebraically and graphically.
Answer: Let us assume that cost of 1 kg apple = x and cost of 1 kg grapes = y. Following equations can be written as per the question.
2x + y = 160 ……..(1)
This equation gives following values for x and y
x | 70, 60, 50, 40 |
y | 20, 40, 60, 80 |
4x + 2y = 300
Or, 2x + y = 150 ……..(2)
This equation will give following values for x and y:
x | 65, 55, 45, 35 |
y | 20, 40, 60, 80 |
The following graph is plotted for the given pair of linear equations.
Since we get parallel lines so there will be no Answer for this pair of linear equations.