Just as we have used cups and beans or chips to learn earlier math concepts, we can use these materials to teach students how to solve one-step algebra equations.
For the purposes of this lesson, a cup represents the variable, white counters represents positive integers and black represents negative integers. After representing the problem with the cup and counters, the goal is to get the cup by itself on one side of the mat by using the following rules:
- A zero-pair is formed by pairing one positive identical counter with one negative counter.
- You can remove or add the same number of identical counters to each side of the equation mat.
- You can remove or add zero-pairs to either side of the equation mat without changing the equation.
For our first teaching problem, we will use the equation x + (-3) = -5. Place 1 cup and 3 negative counters on one side of the mat. Place 5 negative counters on the other side of the mat.
Remove 3 negative counters from each side to get the cup by itself.
The cup on the left side is matched with 2 negative counters. Therefore, x = -2.
Now, let us solve the problem, 2p = -6.
Place 2 cups on one side of the mat. Place 6 negative counters on the other side of the mat.
Separate the counters into 2 equal groups to correspond to the 2 cups.
Each cup on the left is matched with 3 negative counters. Therefore, p = -3.
Lastly, let's solve the equation r - 2 = 3.
Let's change the equation to r + (-2) = 3. Place 1 cup and 2 negative counters on one side. Place 3 positive counters on the other side.
Notice that it is not possible to remove the same kind of counters from each side. Add 2 positive counters to each side.
Group the counters to form zero-pairs. Then, remove all zero-pairs.
The cup on the left is matched with 5 positive counters. Therefore, r = 5.
Students can now use what they have learned to solve equations you give them or they can write their own. They can justify their answer with a sketch in their math journals. As a quiz, they can write a paragraph explaining why zero-pairs can be used to solve an equation such as m + 5 = -8.