Substitution Method

This method mainly involves two steps: 

Step 1: Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.

Step 2: Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved.

Question: Solve the following system of equations: 

7x – 15y = 2

x + 2y = 3

Solution: 

It can be solved according to the steps explained above, 

We pick an equation to represent value of one variable in terms of others, 

Let’s pick, x = 3- 2y (It’s also convenient). 

Substitute the value of x in the other equation. 

7(3-2y) -15y = 2 

⇒ 21 – 14y -15y = 2 

⇒ 19 = 29y 

⇒ y = 19/29

Now, the x value becomes x = 3 – 2(19/29) = 3 – (38/29) = 49/29

This solution can also be checked by substituting both the values in these equations. 

Note: We have substituted the value of one variable by expressing it in terms of other variable to solve the problem. That’s why this method is called substitution.

Sometimes, as in the Example below, we can get statements with no variable. If this statement is true, we can say that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.

Question: Anuj and Rahul bought some stationery items from the shop. Anuj bought 2 pencils and 3 erasers. The cost of 2 pencils and 3 erasers was Rs 9 and Rahul bought 4 pencils and 6 erasers. The cost of 4 pencils and 6 erasers is Rs 18. Find the cost of each pencil and each eraser.

Solution:

The pair of linear equations that is formed from the above description is,

2x + 3y = 9 

4x + 6y = 18

where x is the cost of a pencil and y is the cost of eraser. 

x = [Tex]\frac{9  – 3y}{2}[/Tex]

Putting this value in the other equation. 

[Tex]4(\frac{9 – 3y}{2}) + 6y = 18 \\ 18 – 6y + 6y  = 18 \\ 18 = 18[/Tex]

This statement is true for all values of y. However, we do not get a specific value of y as a solution. Therefore, we cannot obtain a specific value of x. This situation has arisen because both the given equations are the same. Therefore, these equations both have infinitely many solutions. 

When solving a pair of linear equations in two variables, there are several algebraic methods you can use. Here’s a summary of the most common methods:

Let’s imagine a situation, Ankita went to a fair in her village. She wanted to go on rides like the Giant Wheel and play Hoopla (a game in which a ring is thrown on the items kept in a stall, and if the ring covers any object completely, the player gets it).

The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. If each ride on the giant wheel costs Rs 3, and a game of Hoopla costs Rs 4, find out the number of rides she had and how many times she played Hoopla, provided she spent Rs 20.

For solving such kind of problem, the first step is to formulate it in terms of equations. Let x be the number of times Ankita rode on the Giant Wheel and y be the number of times she played Hoopla. The equations become, 

x = 2y, 

3x + 4y = 20.

Can we find the solution to this system of equations? There are several ways of finding solutions, let’s look at them in detail.