Solving Systems of Equations by Substitution

This algebra lesson introduces the technique of solving systems of equations by substitution.

In some word problems, we may need to translate the sentences into more than one equation. If we have two unknown variables then we would need at least two equations to solve the variable. In general, if we have n unknown variables then we would need at least n equations to solve the variable.

In the Substitution Method, we isolate one of the variables in one of the equations and substitute the results in the other equation. We usually try to choose the equation where the coefficient of a variable is 1 and isolate that variable. This is to avoid dealing with fractions whenever possible. If none of the variables has a coefficient of 1 then you may want to consider the Addition Method.

Let’s look at an example.

Example:

3x + 2y = 2      (equation 1)
y + 8 = 3x        (equation 2)

Solution:

Step 1: Try to choose the equation where the coefficient of a variable is 1.

Choose equation 2 and isolate variable y
y = 3x – 8         (equation 3)

Step 2: From equation 3, we know that y is the same as 3x – 8

We can then substitute the variable y in equation 1 with 3x – 8
3x + 2 (3x – 8) = 2

Step 3: Remove brackets using distributive property

3x + 6x – 16 = 2

Step 4: Combine like terms

9x – 16 = 2

Step 5: Isolate variable x

9x = 18

Step 6: Substitute x = 2 into equation 3 to get the value for y

y = 3 (2) – 8
y = 6 – 8 = – 2

Step 7: Check your answer with equation 1

3 (2) + 2 (–2) = 6 – 4 = 2

Answer: x = 2 and y = –2

The following video shows another example of of solving systems of equations by substitution.

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