Pair of Linear Equations in Two Variables

Inconsistent, Consistent and Dependent Pairs of Linear Equations

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

2x – y = – 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 + 3x – y – 3

3x – y = 3 … (2)

Thus, the algebraic representation of the given situation is

2x – y = – 2 … (1)

3x – y = 3 … (2)

Suppose you go to a stationary shop to buy some registers and pens. If you buy two registers and two pens, then you have to pay Rs 30. However, if you buy two registers and four pens, then you have to pay Rs 40.

Now, how can we represent this information algebraically, i.e., in the form of equations?

Let us denote the cost of each register by Rs x and that of each pen by Rs y.

Now, it is given to us that two registers and two pens cost Rs 30.

∴ 2x + 2y = 30

⇒ x + y − 15 = 0 … (1)

We also know that two registers and four pens cost Rs 40.

∴ 2x + 4y = 40

⇒ x + 2y − 20 = 0 … (2)

Thus, we now have two linear equations in two variables from the given situation.

x + y − 15 = 0 … (1)

x + 2y − 20 = 0 … (2)

These two equations are known as a pair of linear equations in two variables. If we try to individually solve each equation, we will be unable to do so. We can arrive at an answer only if we try and solve both these equations together.

The general form of a pair of linear equations in two variables is written as

, where a1, b1, c1 are real numbers, and (i.e., a1, b1 ≠ 0)

, where a2, b2, c2 are real numbers, and (i.e., a2, b2 ≠ 0)

A pair of linear equations represents two straight lines. When their graphs are drawn, there are three possibilities.

(i) The two lines may intersect at one point.

(ii) The two lines may be parallel.

(iii) They may represent the same line, i.e. they may be coincident.

The solution of a pair of linear equations is the point of intersection of their graphs.

Let us understand the above stated statement with the help of the video.

Now, according to this, a pair of dependent equations is represented by coincident lines. Let us verify this graphically with the help of the given video.

Similarly, you can also graphically prove two lines to be inconsistent. Try it out yourself.Limitations of graphical method:

If the large values of variables are involved in the solution of system of linear equations then it is not convenient to use graphical method. If the fractional values of variables are involved in the solution of system of linear equations then it is not possible to obtain accuracy through graphical method. If the angle between two lines is small then the exact point of intersection cannot be obtained through graphical method.Now, let us solve some examples to understand this concept better.

Example 1:

Anshuman and Vik…

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