System of linear equations, also called simultaneous equations, can also be solved using the substitution method. This lesson will show how to solve a pair of linear equations with two unknown variables.

ax + by = c

dx + ey = f

Before you read this lesson, make sure you understand how to solve linear equations.

- Pick an equation and solve either for x or for y. It may be beneficial to pick the equation where it is easy to solve for the variable.
- If you solved for x in step 1, substitute the value of x into the other equation. Otherwise, if you solved for y, substitute the value of y into the other equation.
- If you solved for x in step 1 and substituted the value of x into the other equation, you will end up with a linear equation where the variable is y. Solve for y!
**If on the other hand**, you solved for y in step 1 and substituted the value of y into the other equation, you will end up with a linear equation where the variable is x. Solve for x! - Plug the value of the variable you found in step 3 into one of the original equations in order to find the value of the other variable.

**Example #1**: Solve the following system using the substitution method

x + y = 20

x − y = 10**Step 1**

You have two equations. Pick either the top equation or the bottom equation and solve for either x or y.

Since I am the one solving it, I have decided to choose the equation at the bottom (x − y = 10) and I will solve for x.

x − y = 10

Add y to both sides

x − y + y = 10 + y

x = 10 + y**Step 2**

Since you used the equation at the bottom and solved the equation for the variable x, you will substitute x into the equation on top (x + y = 20)

Using x + y = 20, erase x and write 10 + y since x = 10 + y

We get 10 + y + y = 20

**Step 3**

Solve for y

10 + y + y = 20

10 + 2y = 20

Minus 10 from both sides

10 − 10 + 2y = 20 − 10

2y = 10

Divide both sides by 2

y = 5

**Step 4**

Since you have found the value of the variable y, you can plug its value into either the top equation or the bottom to get x.

Replacing y into x + y = 20 gives x + 5 = 20

Minus 5 from both sides

x + 5 − 5 = 20 − 5

x = 15

The solution to the system is x = 15 and y = 5**Indeed 15 + 5 = 20 and 15 − 5 = 10**

**Example #2**: Solve the following system using the substitution method

3x + y = 10

-4x − 2y = 2**Step 1**

You have two equations. Pick either the first equation (top) or the second equation (bottom) and solve for either x or y.

I have decided to choose the equation on top (3x + y = 10) and I will solve for y.

3x + y = 10

Subtract 3x from both sides

3x − 3x + y = 10 − 3x

y = 10 − 3x**Step 2**

Since you used the equation on top to solve for y, you will substitute y into the equation at the bottom (-4x − 2y = 2)

Using -4x − 2y = 2, erase y and write 10 − 3x keeping in mind that there is a multiplication between 2 and y.

We get -4x − 2×(10 − 3x ) = 2**Step 3**

Solve for x

-4x − 2 ×(10 − 3x ) = 2

-4x − 20 + 6x = 2 (After multiplying -2 by 10 and -2 by -3x)

2x − 20 = 2

Add 20 to both sides

2x − 20 + 20 = 2 + 20

2x = 22

Divide both sides by 2

x = 11

**Step 4**

Now you have x, you can replace its value into either the first equation on top or the second equation at the bottom to get y.

Replacing x into 3x + y = 10 gives 3 × 11 + y = 10

33 + y = 10

Minus 33 from both sides

33 − 33 + y = 10 − 33

y = -23

The solution to the system is x = 11 and y = -23**Indeed,**** 3 × 11 + -23 = 33 + -23 = 10 and -4 × 11 − 2 × -23 = -44 + 46 = 2**

You should have noticed so far that the reason we call this method the substitution method is because after you have solve for a variable in one equation,you substitute the value of that variable into the other equation.

**Example #3**: Solve the following system using the substitution method

2x + y = 8

2x + y = 8

**Step 1**

Pick the equation on top and solve for y.

2x + y = 8

2x - 2x + y = 8 - 2x

y = 8 - 2x

**Step 2**

Substitute the value of y in the equation at the bottom.

2x + 8 - 2x = 8

8 = 8

Notice that instead of finding a value for x, we end up with a true statement (8 = 8)

When this happens, it means that the system has an infinite number of solutions. Any ordered pair (x,y) that solves the system is a solution. Find ordered pairs by choosing anything you want for x and then solve for y.

For example, if I choose 1 for x, then 2(1) + y = 8

2 + y = 8

y = 6

A solution is (1, 6). You can find an infinite amount if you keep choosing arbitrary value for x.

If you solve a system by substitution and you end up with a true statement, it means that the system has an infinite number of solutions.

**Example #4**: Solve the following system using the substitution method

2x + y = 4

2x + y = 8

**Step 1**

Pick the equation on top and solve for y.

2x + y = 4

2x - 2x + y = 4 - 2x

y = 4 - 2x

**Step 2**

Substitute the value of y in the equation at the bottom.

2x + 4 - 2x = 8

4 = 8

Notice that instead of finding a value for x, we end up with a false statement (4 = 8)

If you solve a system by substitution and you end up with a false statement, it means that the system has no solution.