Tape diagrams

Tape diagrams are a visual representation of quantities that use length to show equality. Equal sides of an equation have the same lengths and are stacked on one another.

For example, our equation "$x=5$" is pictured as a tape diagram above. It’s important to understand that equations and tape diagrams are merely different representations of the same thing (they are “equal” in a sense). "$x=5$" is one way to say “x is the same as 5,” and so is our tape diagram.

For this tape diagram, it’s arbitrary how we arranged it. We could have just as easily put the $5$ on top and the $x$ on the bottom, just as we could instead write the equation "$5=x$" to convey the same thing.

Tape diagrams are most useful to help visualize equations that use basic arithmetic ($+$, $-$, $\times$, $\div$). We will discuss extensions of this model for more complicated equations as we delve further.

Examples of equalities represented as tape diagrams:

Note that for (c), we can visualize the subtraction of $5$ from $3x$ by placing it below the $3x$ to show that by removing 5 from the length of 3x, we arrive at a length of 16. Algebraically, this is the same as adding 5 to both sides of the equation.

Negative signs

With negative numbers, it’s fairly straightforward to represent them in a tape diagram by placing them on the other side of the diagram, so to speak. We can visualize subtracting from a quantity by “cutting off” that much length. This aligns with the algebraic idea of using zero-pairs to balance an equation.

However, with equations like "$3x = -18$" the visualization starts to break down. We can fix this by relativizing our tape diagrams to 0. Then, we anchor our variables to our constant quantity, which may lie on either side of 0.

Looking at the diagram to the left, it should be more visually obvious that each $x$ should account for taking us 6 to the left (i.e., $x$ takes on a value of $-6$).

The diagram to the right is a representation of "$-5x = 30$". Using a tape diagram can help us visually deduce what the value of $x$ must be, but just as in solving equations algebraically, we must be especially aware of potential sign errors and misconceptions. Note that each box above the “30” is "$-x$". So, while we can easily see that each box has a length of 6, we must be careful to see that $x$ will equal $-6$, not $6$.

As you can see, introducing negatives pushes tape diagrams to their limits, so to tackle harder equations, we’ll want to build a robust algebraic process, which we’ll get to shortly.

Tape diagrams for absolute value equations

Extending our tape diagram model of equality to accommodate negative values gives us a nice visual way to solve absolute value equations.

So, how do we take something like $|x + 2| = 6$ and represent it visually?

First, what does this equation represent? It says that if I take the absolute value of some quantity ($x + 2$, in this case) it should equal $6$.

We’ve previously discussed how absolute value can be thought of as “distance from zero,” and that distance has no sign (it is never negative). So, if we relativize to $0$ and ask, “where on the number line are we 6 away from 0?” we end up with two answers (both at $6$ and $-6$). Let’s visualize this:

Moving from tape diagrams to a robust algebraic process

I like tape diagrams as a way to build a foundational, reasoning-based approach to solving for a variable or isolating a term in an equation. Look at the equations and their visual representation below, and try to reason what value $x$ should take.

When you look at the diagrams, hopefully you’re able to discern that for (i), $x$ should be 7, and for (ii), $x$ should be 6. What did your reasoning look like?

Example reasoning:

Looking at (i), we can notice that there are 2 x’s on the top and bottom that both account for the same length on each tape. So, we can use the leftover x on the top and the rest of the diagram to reason what $x$ must be.

Since there is a length of $5$ accounted for on the top tape to the right of $x$, and the total length must be $12$ to match the bottom, then $x$ must have a length of $7$ ($7+5=12$, making the two segments match in length). Therefore, the solution to (i) is $[x = 7]$.

Similarly, for (ii), we can see that each side (referring to the top and bottom tapes, respectively) has 3 x’s accounting for part of its length. That means that the remaining segments must each have equal length.

Algebraically, this reduces our equation to $3x+10=28$. Think about what algebraic step would get us to that point from our original equation, and how that algebraic step relates to the reasoning we just performed by “removing” 3 x’s from our tape diagram.

Our next step would be to determine what length the 3 remaining x’s on the top must occupy. Since there is a segment of $10$ accompanying them, and the total must equal a length of $28$, we can deduce that the 3 x’s must occupy a total length of 18.

Then, we ask ourselves the question, “Can I divide 18 into 3 equal parts?” and the answer is, of course, “Yes!” And we’ve reasoned that $x$ must be $6$.

Therefore, the solution to (ii) is $[x = 6]$.

Each step of our reasoning can be represented as an algebraic step as well. Let’s walk through the algebraic process of solving equation (i):

Solving equation (ii):

Try to find the parallels between the algebraic solution and the solution found by visual reasoning.