# How to Solve Systems of Equations by Substitution

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2. Solving Systems of Equations by Substitution

### The Substitution Method

Substitution is the quickest method of solving a system of two equations in two variables. The method can also be used to find the solution of a system of three or more equations in three or more variables, but it takes longer.

In this lesson, we are going to talk you through solving a system of 2 linear equations in 2 variables.

### Solving Systems of Equations by Substitution

The substitution method involves three steps. They are:

Rearrange an equation to isolate one of the variables on one side.

Substitute the expression so obtained into the other equation to solve for the other variable.

Plug the value back into one of the equations to solve for the variable initially isolated.

Let's understand the steps with this linear system.

y = x + 6

5x + 2y = 19

In the above system, the variable y is isolated. So, the step 1 is already accomplished. Now, let’s substitute the expression for y in the second equation.

5x + 2(x + 6) = 19

Applying the distributive property, we have:

5x + 2x + 12 = 19

Combining like terms, we get:

7x + 12 = 19

Subtracting 12 from both sides, we have:

7x + 12 –12 = 19 –12

7x = 7

Dividing both sides by 7, we get:

7x7 = 77

⇨ x = 1

Now the last step! We need to plug this value of x back in one of the equations. Let’s plug it in y = x + 6.

y = 1 + 6

⇨ y = 7

Thus, the solution of the linear system is (1, 7).

Let’s take a look at another example.

Example

Solve 4x – 6y = –16 and 8x + 2y = 24 using the substitution method.

Let’s choose the second equation this time. Solving the second equation for y, we get:

8x + 2y = 24

2y = –8x + 24

y = –4x + 12

Now substituting y in the first equation, we have:

4x – 6(–4x + 12) = –16

4x + 24x – 72 = –16

28x = –16 + 72

28x = 56

⇨ x = 2

We can put this x-value back into any of the given equations and solve for y.

But,do you remember?

We have rearranged the equation 2 and have an expression for y all set!

y = –4x + 12

Now, all we need to do is just substitute the value of x into this.

y = –4x + 12

y = –4(2) + 12

y = –8 + 12

⇨ y = 4

Thus, the solution is (x, y) = (2, 4).

You can cross-check if your solution is right by substituting the x and y values in the equations.
Plugging (2, 4) in equation 1, you get:
4(2) – 6(4) = –16
8 – 24 = –16
–16 = –16

Plugging (2, 4) in equation 2, you get:
8(2) + 2(4) = 24
16 + 8 = 24
24 = 24

### Which Variable to Isolate When Solving a System with Substitution

It doesn’t matter which equation you choose or which variable you solve first; the solution of the system remains the same. Let’s take an example to illustrate this fact.

Solve this system by substitution.

–x + y = –4

4x – 3y = 10

We’ve shown you the step-by-step solution starting with x as well as y.

Isolate x First

Isolating x from equation 1, we have:

⇨ –x + y – y = –4 – y
[Subtracting y from both sides]

⇨ –x = –4 – y [Combining like terms]

⇨ x = 4 + y [Multiplying throughout by –1]

Now, substituting x in equation 2, we have:

⇨ 4(4 + y) – 3y = 10

⇨ 4(4) + 4y – 3y = 10
[Applying distributive property]

⇨ 16 + 4y – 3y = 10

⇨ 16 + y = 10 [Combining like terms]

⇨ 16 + y – 16 = 10 – 16
[Subtracting 16 from both sides]

⇨ y = –6 [simplifying]

Plugging back y in x = 4 + y, we get:

⇨ x = 4 – 6

⇨ x = –2

Therefore, the solution set is (–2, –6).

Isolate y First

Isolating y from equation 1, we have:

⇨ –x + y + x = –4 + x [Adding x to both sides]

⇨ y = –4 + x [Combining like terms]

Now, substituting y in equation 2, we have:

⇨ 4x – 3(–4 + x) = 10

⇨ 4x – 3(–4) – 3x = 10
[Applying distributive property]

⇨ 4x + 12 – 3x = 10

⇨ x + 12 = 10 [Combining like terms]

⇨ x + 12 – 12 = 10 – 12
[Subtracting 12 from both sides]

⇨ x = –2 [simplifying]

Plugging back x in y = –4 + x, we get:

⇨ y = –4 – 2

⇨ y = –6

Therefore, the solution set is (–2, –6).

As you can see, the solution in both the cases is the same. So, just get started with a step that’s most convenient and simple.

### The Gist of What We Learned So Far!

• Solving a system of equations in two variables using substitution is easy and quick!

• To solve by substitution, you need to follow a three-step process that involves isolation of a variable, substitution of the expression, and re-substitution of the value.

• No matter which equation you choose and which variable you solve first; the solution of the system of equations stays the same. Refine your practice with our free printable Solving Systems of Linear Equations worksheets!

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