In the previous sections we looked at the addition method for solving systems of equations in two unknowns, but that is not always the easiest method. Let's take a look at an example where a method called the substitution method may be easier.

Example #1

x + y = 10
y = 3x + 2

If we were going to use the addition method to solve this, we would need to rearrange the second equation to get it into the nice format that we're accustomed to. But there's something very different we can do here. Notice that the second equation is solved for y. This doesn't mean we know what y equals; it just means that y is all by itself on one side of the equation, and everything else is on the other side of the equation. Since y = 3x + 2, we could replace y with (3x + 2) in the first equation:

x + y = 10
x + (3x + 2) = 10

Now we can just combine like terms and solve for x:

4x + 2 = 10
4x = 8
x = 2

At this point, we can plug the x into one of our equations and solve for y. But since one equation is already solved for y, it's a no-brainer; we're going to plug x into that equation:

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

Our solution is x = 2; y = 8

Example #2

x + 2y = 13
y = 2x - 1

Since y = 2x - 1, why don't we replace y with (2x - 1) in the first equation?

x + 2y = 13
x + 2(2x - 1) = 13
x + 4x - 2 = 13
5x - 2 = 13
5x = 15
x = 3

Plug this into the second equation, and we get:

y = 2x - 1
y = 2(3) - 1
y = 6 - 1 = 5

So the answer is: x = 3; y = 5

Example #3

2x - 3y = 16
y = 2x - 12

Pay close attention here, because the coefficient of y in the first equation is negative, which isn't a problem, but it means we're going to be distributing a negative through our parentheses, which will change the sign of every term inside!

2x - 3y = 16
2x - 3(2x - 12) = 16
2x - 6x + 36 = 16
-4x  = -20
x = 5

Then we can plug this into the second equation:

y = 2x - 12
y = 2(5) - 12 = 10 - 12 = -2

Example #4

x + 2y = 15
2x - y = 0

In this case, neither equation is solved for a variable. For many students (and teachers) the preferred choice in this situation is to use the addition method instead of the substitution method. However, it's good practice to use the substitution method, just to make sure you can properly rearrange the terms to make it work. Once you've had the practice, most teachers and textbooks will tell you that you can choose which method you prefer.

Our choices are: solve the first equation for x, solve the first equation for y, solve the second equation for x, or solve the second equation for y. My choice always is: if there's a variable that doesn't have a coefficient, that's what I'm going to solve for, because it avoids having to do a division (and potentially ending up with fractions!). So here I'll solve the second equation for y by adding y to both sides of the equation:

2x - y =  0
   + y  + y
_________
2x     =  y

Now, since y = 2x, we can substitute that into the first equation:

x + 2y = 15
x + 2(2x) = 15
x + 4x = 15
5x = 15
x = 3

And, of course, we plug that back into the second equation to obtain y = 6.

In the exercises below, ask your teacher if they want you to use substitution for all of them, of they'll let you choose your method.

1.

Under what circumstances is the substitution method most useful?

2.

If you have to solve for a variable first, in order to use the substitution method, why is it easiest to solve for a variable without a coefficient?

3.

Solve the system: 2x + y = 10; y = x + 1

4.

Solve the system: 3x - y = 14; x = 2y + 3

5.

Solve the system: x + y = 2; y - x = 0

6.

Solve the system: 3x + y = 1; y = 2x + 6

7.

Solve the system: x - y = -1; x + y = 5

8.

Solve the system: x + y = 22; 2x + 3y = 56

9.

Solve the system: x = 2(y - 8) - 1; x + y = 13

10.

Solve the system: 2(x + 1) = y + 2; y - x = 1