Class 10 Maths

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.

Solution: 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;

`x – 7 = 7(y – 7)`
Or, `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;

`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:

linear equations exercise solution

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.

Solution: Let us assume that price of one bat = x and price of one ball = y. Following equations can be written as per the question:

`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

`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.

linear equations exercise solution

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.

Solution: 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.

linear equations exercise solution

Since we get parallel lines so there will be no solution for this pair of linear equations.