Chapter 3 · Algebraic foundations
Model and Solve Equations with Algebra Tiles
See an equation as tiles on a balance, then solve it with moves that keep both pans equal: the same tile on both pans, zero pairs and equal groups.
You can already solve two-step equations by doing the same inverse operation to both sides. Algebra tiles show those steps as moves you can see. In this lesson, stands for an unknown number, and it may be any real number.
Each tile is labelled with its value. A unit tile is a small square: a positive unit tile is worth and a negative unit tile is worth . A variable tile is longer: a positive variable tile is worth and a negative variable tile is worth .
To model an equation, put tiles for its left side on the left pan of a balance and tiles for its right side on the right pan. A pan is worth the sum of its tiles, so a pan holding one tile and four tiles is worth , and an empty pan is worth . The equal sign says that the two pans have the same value.
A tile and a tile on the same pan form a zero pair, because . A tile and a tile on the same pan also form a zero pair, because for every value of .
Key concept
Moves that keep the pans equal
Each move below keeps the two pans equal in value, so each new board is an equivalent equation: one with exactly the same solution.
- Same tile on both pans. Add the same tile to both pans, or remove the same tile from both pans. This adds or subtracts the same amount on both sides.
- Zero pair on one pan. Remove a zero pair from one pan. It is worth , so that pan keeps its value.
- Equal groups. When the tiles on both pans sort into the same number of equal groups, at least , remove one group from each pan until one group is left. This divides both sides by the number of groups, which is never .
Each equation in this lesson has exactly one solution, a whole number, so the moves can end with one tile alone on a pan. The other pan then shows the solution.
Algebra tiles on a balance, in 4 boards. Board 1 models 3x - 2 = 4. Left pan: 2 negative unit tiles, 3 positive variable tiles. Right pan: 4 positive unit tiles. Board 2: 2 positive unit tiles are added to both pans. Left pan: 2 positive unit tiles, 2 negative unit tiles, 3 positive variable tiles. Right pan: 6 positive unit tiles. This shows 3x + 2 - 2 = 6. Board 3: 2 zero pairs of unit tiles are removed from the left pan. Left pan: 3 positive variable tiles. Right pan: 6 positive unit tiles. This shows 3x = 6. Board 4: The board is 3 equal groups, so one group is removed from each pan. The board is 2 equal groups, so one group is removed from each pan. Left pan: 1 positive variable tile. Right pan: 2 positive unit tiles. This shows x = 2. So x = 2.
Worked example
Use algebra tiles to solve .
Model the equation. The left side, 3x - 2, becomes three positive variable tiles and two negative unit tiles on the left pan. The right side, 4, becomes four positive unit tiles on the right pan. This is board 1.
The two negative unit tiles need partners to form zero pairs, so add two positive unit tiles to both pans. Adding the same tiles to both pans adds 2 to both sides, so the pans stay equal. The right pan now holds 4 + 2 = 6 positive unit tiles.
The left pan now holds two zero pairs. Remove them. Each pair is worth 0, so the left pan keeps its value.
The tiles now sort into 3 equal groups, each with one positive variable tile on the left pan and two positive unit tiles on the right pan. Remove one group from each pan, then another, so one group is left. Keeping 1 of 3 equal groups divides both sides by 3, which is not 0.
Check in the original equation by replacing x with 2. Three times 2 is 6, and 6 minus 2 is 4, which matches the right side.
Answer
The solution is , because is true.
The model of an equation can also hold a negative variable tile, as in . Your goal is one tile alone, and no move turns a tile into a tile. Instead, add a tile to both pans. On the pan with the tile, the new tile makes a zero pair. On the other pan, it is the you are solving for. So may finish on the right pan, and an equation such as says the same thing as .
Algebra tiles on a balance, in 4 boards. Board 1 models -x + 6 = 2. Left pan: 6 positive unit tiles, 1 negative variable tile. Right pan: 2 positive unit tiles. Board 2: 1 positive variable tile is added to both pans. Left pan: 6 positive unit tiles, 1 positive variable tile, 1 negative variable tile. Right pan: 2 positive unit tiles, 1 positive variable tile. This shows x - x + 6 = x + 2. Board 3: 1 zero pair of variable tiles is removed from the left pan. Left pan: 6 positive unit tiles. Right pan: 2 positive unit tiles, 1 positive variable tile. This shows 6 = x + 2. Board 4: 2 positive unit tiles are removed from both pans. Left pan: 4 positive unit tiles. Right pan: 1 positive variable tile. This shows 4 = x. So x = 4.
Worked example
Use algebra tiles to solve .
Model the equation: one negative variable tile and six positive unit tiles on the left pan, and two positive unit tiles on the right pan.
Add one positive variable tile to both pans. This adds x to both sides, so the pans stay equal.
On the left pan, the positive and negative variable tiles form a zero pair. It is worth 0 for every value of x, so remove it.
Remove two positive unit tiles from both pans, which subtracts 2 from both sides. One positive variable tile now stands alone on the right pan, and the left pan holds 6 - 2 = 4 positive unit tiles.
Check in the original equation by replacing x with 4.
Answer
The solution is , because is true.
Sometimes both pans hold the same kind of tile. In the model of , each pan holds a tile. Removing a tile that both pans share is a same-tile move, and it leaves fewer tiles that need partners. This example uses every kind of move.
Algebra tiles on a balance, in 5 boards. Board 1 models 2x - 3 = -1. Left pan: 3 negative unit tiles, 2 positive variable tiles. Right pan: 1 negative unit tile. Board 2: 1 negative unit tile is removed from both pans. Left pan: 2 negative unit tiles, 2 positive variable tiles. Right pan: empty. This shows 2x - 2 = 0. Board 3: 2 positive unit tiles are added to both pans. Left pan: 2 positive unit tiles, 2 negative unit tiles, 2 positive variable tiles. Right pan: 2 positive unit tiles. This shows 2x + 2 - 2 = 2. Board 4: 2 zero pairs of unit tiles are removed from the left pan. Left pan: 2 positive variable tiles. Right pan: 2 positive unit tiles. This shows 2x = 2. Board 5: The board is 2 equal groups, so one group is removed from each pan. Left pan: 1 positive variable tile. Right pan: 1 positive unit tile. This shows x = 1. So x = 1.
Worked example
Use algebra tiles to solve .
Model the equation: two positive variable tiles and three negative unit tiles on the left pan, and one negative unit tile on the right pan.
Each pan holds a negative unit tile, so remove one negative unit tile from both pans. Taking away a tile worth -1 from both pans adds 1 to both sides, so the pans stay equal. The right pan is now empty, so it is worth 0.
Add two positive unit tiles to both pans, so each negative unit tile on the left has a partner. This adds 2 to both sides.
Remove the two zero pairs from the left pan. Each is worth 0, so the left pan keeps its value.
The tiles sort into 2 equal groups, each with one positive variable tile on the left pan and one positive unit tile on the right pan. Remove one group from each pan. Keeping 1 of 2 equal groups divides both sides by 2, which is not 0.
Check in the original equation by replacing x with 1.
Answer
The solution is , because is true.
Takeaway. A tile model shows an equation as a balance: each pan holds one side, and the two pans have the same value. Adding or removing the same tile on both pans, removing a zero pair from one pan and removing equal groups all keep the pans equal, so every board has the same solution. Write each board's equation as you go. When one tile stands alone, the other pan shows the solution; check it in the original equation.
Next, you will write and solve equations for word problems.