# How to Solve Systems of Equations by Substitution

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- Solving Systems of Equations by Substitution

### Overview of 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:

*7x***7** = *7***7**

⇨ 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).

Check Your Solution!

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!