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

**Substitution method****Elimination method****Cross-multiplication method**

In this article, we will focus mainly on the Substitution method.

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.

Let us understand the above theory of substitution using substitution method examples.

## Substitution method for solving a pair of linear equations:

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 other 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.

Let us consider another example, in which we will learn about the substitution method.

**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 Substitution 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!