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Pairs of Equations

Solution of Linear Equations That Contain Linear Expressions on one Side and Numbers on the Other Side

We come across many situations in real life when it is easy to find the solution, if we express them mathematically.

Let us see such a situation.

The coach of the school cricket team buys 5 bats and 20 leather balls for Rs 3500. After some time, some more boys joined the team for practice, so he buys another 4 bats and 15 balls for Rs 2750. Suppose that the price of bat and ball does not change in the time period.

Can we express this situation mathematically to find out the individual prices of a ball and a bat?

Let the price of a bat be Rs x and that of a ball be Rs y.

It is given that 5 bats and 20 balls cost Rs 3500.

∴ Cost of 5 bats = 5x

And cost of 20 balls = 20y

⇒ Cost of 5 bats and 20 balls = 5x + 20y

⇒ 5x + 20y = 3500

Similarly, it is also given that 4 bats and 15 balls cost Rs 2750.

⇒ 4x + 15y = 2750

Therefore, the algebraic representation of the given situation is

5x + 20y = 3500 … (1)

4x + 15y = 2750 … (2)

After solving these equations, we can find out the individual prices of the ball and the bat.

Let us see some more examples.

Example 1:

Aman and Yash each have certain number of chocolates. Aman says to Yash, if you give me 10 of your chocolates, I will have twice the number of chocolates left with you. Yash replies, if you give me 10 of your chocolates, I will have the same number of chocolates as left with you. Write this situation mathematically?

Solution:

Suppose Aman has x number of chocolates and Yash has y number of chocolates.

According to the first condition, Yash gives 10 chocolates to Aman so that Aman has twice the number of chocolates than what Yash has.

x + 10 = 2(y – 10)

According to the second condition, Aman gives 10 chocolates to Yash such that both have equal number of chocolates.

y +10 = x – 10

Thus, the algebraic representation of the given situation is

x + 10 = 2(y – 10) … (1)

y +10 = x – 10 … (2)

Example 2:

Cadets in the military academy are made to stand in rows. If one cadet is extra in each row, then there would be 2 rows less. If one cadet is less in each row, then there would be 3 more rows. Express the given situation mathematically?

Solution:

Let the number of cadets in each row be x and the number of rows be y.

Total number of cadets = number of rows number of cadets in each row

It is given to us that when one cadet is extra in each row, there are 2 rows less.

xy = (y – 2) (x + 1)

xy = xy – 2x + y – 2

2xy = – 2 … (1)

It is also given to us that if one cadet is less in each row, then there are 3 more rows.

xy = (y + 3) (x – 1)

xy = xy + 3xy – 3

3xy = 3 … (2)

Thus, the algebraic representation of the given situation is

2xy = – 2 … (1)

3xy = 3 … (2)

In our daily life, we come across many types of linear equations. Let us take an example.

Suppose, there are 30 students in a class and they are collecting money to donate for a charity cause. If each of the students pays the same amount and their teacher pays Rs 100, then the total amount collected will be Rs 1000.

Now can we find out the amount that each student contributed?

The given video will help us to find out the answer.

Let us now solve some more equations which contain expressions on one side and numbers on the other side.

Example 1:

Solve the following linear equations.

I. II. 12x − 13 = 131

Solution:

I. On adding to both the sides, we obtain On multiplying both the sides with 2, we obtain

x = 10

We can also check the answer by putting the value of x on LHS of the given equation.

Check: = R.H.S.

II. 12x − 13 = 131

On adding 13 to both the sides, we obtain

12x − 13 + 13 = 131 + 13

⇒ …

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