Substitution Method - Definition, Steps, Examples

What is Substitution Method?

What is substitution method? It is a commonly asked question among students when learning about algebraic methods of solving linear equations. The substitution method is an easy method to solve a system of linear equations in two variables. In this method, we first find the value of one variable from one equation and then put, or substitute, that value into the second equation. This helps us find the value of the other variable.

Substitution method is very useful when one of the variables in an equation is already simple, like having a coefficient of 1. It is an effective technique for solving equations that makes solving math problems faster and easier. So, keep reading to learn more about the substitution method.

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How to Solve Systems of Equations by Substitution Method?

The substitution method helps us solve two equations with two unknown numbers step by step. Here’s how you solve systems of equations by substitution method:

• Look at the equations carefully: First, check the given equations. If there are brackets, etc., simplify them first, so it becomes easy to solve the rest.

• Solve one equation for one variable: Pick one equation and make one variable (x or y) the subject. It is better to choose the variable that is easier to handle. For example, if the equation is x + y = 5, it is easy to write x = 5 – y.

• Substitute in the second equation: Take the value you found for the variable and put it in the other equation in place of that variable. This gives you a new equation with only one variable.

• Solve the new equation: Now do the math to find the value of the remaining variable. For example, after substituting, you may get 3y – 2 = 4, then solve to find y.

• Find the first variable: Finally, put the value you found back into any of the original equations to get the value of the first variable.

To understand how to solve equations by substitution method more clearly, let’s take an example and solve it step by step.

Ques. Solve the system of equations by substitution method:

1. x + y = 7

2. x – y = 3

Step 1: Solve one equation for a variable: From equation (1), we can write x = 7 – y

Step 2: Substitute in the second equation: Now, put x = 7 – y into equation (2): (7 – y) – y = 3

Step 3: Simplify and solve for y: 7 – y – y = 3

• 7 – 2y = 3

• -2y = 3 – 7

• -2y = -4

• y = 2

Step 4: Find x: Now, put y = 2 into x = 7 – y: x = 7 – 2.
x = 5

Therefore, the final answer is: x = 5, y = 2.

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Difference Between Elimination and Substitution Method

While both the substitution method and the elimination method are methods of solving linear equations with two variables, they differ from each other. Let's understand the difference between elimination and substitution methods:

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Substitution Method Examples With Solutions

Here are some substitution method examples to help you understand how to solve systems of linear equations step by step:

Example 1: Solve the system of equations by substitution method: 4a − 3b = 13 and 2a + b = 7.

Solution: The given equations are:

1. 4a − 3b = 13

2. 2a + b = 7

Step 1: Solve equation (2) for b:
b = 7 − 2a

Step 2: Substitute b = 7 − 2a into equation (1):
4a − 3(7 − 2a) = 13
4a − 21 + 6a = 13
10a − 21 = 13
10a = 34
a = 34 / 10 = 17 / 5

Step 3: Find b by putting a = 17/5 in equation (2):
2 × 17/5 + b = 7
34/5 + b = 7
b = 7 − 34/5 = 35/5 − 34/5 = 1/5

Hence, the final answers are: a = 17/5 and b = 1/5.

Example 2: Two numbers have a sum of 30 and a difference of 12. Find the numbers.

Solution:

Let the two numbers be x and y (x > y). We have:
x + y = 30 … (1)
x − y = 12 … (2)

Step 1: Solve equation (1) for x:
x = 30 − y

Step 2: Substitute x = 30 − y into equation (2):
(30 − y) − y = 12
30 − 2y = 12
−2y = 12 − 30
−2y = −18
y = 9

Step 3: Find x:
x = 30 − 9 = 21

Hence, the two numbers are 21 and 9.

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Example 3: Solve the system:
x + 2y − 4 = 0
2x − y + 5 = 0

Solution:

Step 1: Solve the first equation for x:
x = 4 − 2y

Step 2: Substitute x = 4 − 2y into the second equation:
2(4 − 2y) − y + 5 = 0
8 − 4y − y + 5 = 0
−5y + 13 = 0
−5y = −13
y = 13 / 5

Step 3: Find x:
x = 4 − 2 × 13/5 = 4 − 26/5 = 20/5 − 26/5 = −6/5

Therefore, the answer is = −6/5, y = 13/5

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