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)

Linear equations in two variables are equations where we have two variables. We require two equations to find the solution of linear equations in two variables. Go through the following video to understand the basic concept of linear equations in two variables.

Now the given video will help you to frame equations for a given word problem.

Remember: To solve a linear equation in one variable, only one equation is required. To solve linear equations in two variables, two linear equations in the same two variables are required. Likewise, we can say that to solve linear equations with n number of variables, n numbers of linear equations in the same n number of variables are required.

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

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

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

We know that the solution of a linear equation must satisfy the equation. Conversely, we can say that the value of the variable in the equation, which satisfies the equation, is the solution of the equation.

In case of a pair of linear equations, the values of x and y, which satisfy both the equations, are the solutions of the equation.

Geometrically, a linear equation represents a straight line. Every solution of an equation is a point on the line represented by the equation.

Likewise, a pair of linear equations represents two straight lines. When their graphs are drawn, there are three possibilities. Let us go through the video to understand the various possibilities.

Let us now consider some more examples to check whether the pair of equations is consistent, inconsistent, or dependent.

Example 1:

Find whether the given pairs of linear equations are consistent or inconsistent?

(a) 5x + 2y = 2

20x + 8y = 1

(b) 4x + y = 8

7x – 2y = 1

(c) 5x + 6y = 9

10x + 12y = 18

Solution:

(a) The solution is explained with the help of the following video. Let us go through the video to understand it well.

(b)4 x + y = 8

7x – 2y = 1

Here, a1 = 4 b1 = 1 c1 = –8 a2 = 7 b2 = –2 c2 = –1Therefore, according to the above rule, this system of equations has a unique solution. It is represented by intersecting lines.

(c) 5x + 6y = 9

10x + 12y = 18

Here, a1 = 5 b1 = 6 c1 = –9 a2 = 10 b2 = 12 c2 = –18Therefore, according to the above rule, this system of equations has infinite solutions. It is represented by coincident lines.

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