The solution of the simultaneous linear equation can be divided into two broad categories, Graphical Method, and Algebraic method.

**Algebraic Method**

An Algebraic method is a collection of several methods, which are used to solve a pair of the linear equation that includes two variables. Generally, the algebraic method can be sub-divided into three categories:

- Substitution method
- Elimination method
- Cross-multiplication method

**Graphical Method**

The graphical method is also known as the geometric Method used to solve the system of the linear equation. In this method, the equations are designed based on the objective function and constraints. To solve the system of linear equations, this method has undergone different steps to obtain the solutions.

In this article, we will focus mainly on the first algebraic method called ” **Substitution Method**” in detail.

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## Substitution Method Definition

The** substitution method** is the algebraic method to solve simultaneous linear equations. As the word says, in this method, the value of one variable from one equation is substituted in the other equation. In this way, a pair of the linear equation gets transformed into one linear equation with only one variable, which can then easily be solved.

## Substitution Method Steps

For instance, the system of two equations with two unknown values, the solution can be obtained by using the below steps. Here, the list of steps is provided to solve the linear equation. They are

- Simplify the given equation by expanding the parenthesis
- Solve one of the equations for either x or y
- substitute the step 1 solution in the other equation
- Now solve the new equation obtained using elementary arithmetic operations
- Finally, solve the equation to find the value of the second variable

## Solving Linear Equations by Substitution Method

Let us understand the above theory of substitution using substitution method example

Consider the equations: 2x + 3y = 9 and x – y = 3

For solving simultaneous equations,

Let, 2x + 3y = 9â€¦â€¦..(1) and x – y = 3 â€¦â€¦..(2)

From Equation (2) we get, y = x – 3â€¦â€¦â€¦â€¦â€¦(3)

Now, as we understood above that in the** substitution method**, we find the value of one variable in terms of others and then substitute back.

Now, we know that y = x – 3

Substituting the value of y in equation (1), we get

2x + 3y = 9

â‡’ 2x + 3(x – 3) = 9

â‡’ 2x + 3x – 9 = 9

â‡’ 5x = 18

â‡’ x = \( \frac {18}{5}\)

Now, the value of y can be found out using equation (3)

So, y = x – 3

â‡’ y = \( \frac {18}{5}\) Â – 3

â‡’ y =Â \( \frac {3}{5}\)

Hence the solution of simultaneous equation will be: x = \( \frac {18}{5}\)Â Â and y =Â \( \frac {3}{5}\)

In this way, we can find out the value of the unknowns x and y using substitution method.

### Substitution Method Problems

**Example:**

Solve the pair of linear equations: 4x + 6y = 10 and 2x – 3y = 8 using Substitution method.

*Solution: Â * Â

4x + 6y = 10 ………….(i)

2x – 3y = 8Â ……………(ii)

Finding the value of y in terms of x from equation (1), we get-

4x + 6y = 10

â‡’ 6y = 10 – 4x

â‡’ y = \( \frac {10-4x}{6} \) â€¦â€¦â€¦â€¦â€¦â€¦.(3)

Using this method, substituting the value of y in equation (2), we get-

\( \large 2x – 3 \left( \frac{10~-~4x}{6} \right) \) = 10

â‡’ 2x – 5 + 2x = 10

â‡’ 4x = 15

â‡’ x = \( \frac {15}{4} \)

Finding the value of y, substitute the value of x in equation (3), we get-

y = \(\large \frac {10~-~4*\left( \frac {15}{4} \right)}{6}\)

â‡’ y = \( \frac {10~-~15}{6}\)

â‡’ y = \( \frac {-5}{6} \)

Hence the value of y is \( – \frac 56 \) Â and x is \( \frac {15}{4} \)

Substitution method is generally used for solving simultaneous equations, which are relatively easy. There are direct methods like **cross-multiplication methods** which can directly give you the value of the unknowns, but for simple equations, not involving hectic calculations, this method can be preferred over other algebraic methods- Elimination method and cross-multiplication method.

If the pair of linear equations has no solution, then after substitution you wonâ€™t get the same value of LHS and RHS. In the case of infinite solutions, both sides of the equation will be equal to the same constant.

You will get a unique solution only when you get a proper value of the unknown after substitution.

To learn about other methods to solve linear equations in two variables,Â download BYJUâ€™s-the learning app. Happy Learning!