Simplifying “3-stack” and “4-stack” fractions

By Tutor GuyNo Comments

 

I had a physics student a number of years ago who worked a complicated problem and ended up with the following (I’ve changed the actual numbers to make our work here easier to follow):

\dfrac{(2)(3)}{4}=\dfrac{(5)(6)(7)B}{(8)(9)}

Of course, his next step was to solve for B. It should be clear to you, as it was to him, that he needed to multiply both sides by 8 and 9 and divide by 5, 6 and 7. That is what he did. But this is how he wrote the solution:

B= \dfrac{9 \cdot \left( \dfrac{ \left(\dfrac{ 8 \cdot \left( \dfrac{(2)(3)}{4} \right)}{5} \right)}{6} \right)}{7}

His expression was totally correct, and he found the correct value of B, but he made the problem so much harder than he needed to. He created a fraction with five different stacks in it and he needed all those parentheses to keep track of which number was a numerator and which was a denominator.

This is an extreme case of what I see so many students do: they create “3-stack” and “4-stack” fractions all the time when they are simplifying problems. Let’s look at two examples. I’ll solve them with “bad” solutions and then with much smarter solutions (I don’t simplify the answers here because I want to focus on the first step of the solution):

Example 1. Solve:

3x= \dfrac{2}{5}

Bad solution:

x= \dfrac{\dfrac{2}{5}}{3}

 

Smarter solution:

x= \dfrac{2}{5 \cdot 3}

Example 2. Solve:

\dfrac{4}{7}x=\dfrac{2}{5}

 Bad solution:

x= \dfrac{\dfrac{2}{5}}{\dfrac{4}{7}}

 Smarter solution:

x= \dfrac{2 \cdot 7}{5 \cdot 4}

You can see that the smarter solutions are simpler to read and easier to simplify.

It is easy to train yourself to write answers as “2-stack” fractions if you remember one simple rule:

Respect the vinculum.

Um, respect the what?!?

The vinculum. When you write a fraction, the horizontal line that separates the numerator from the denominator is called the vinculum. No one ever talks about it, but it’s a very powerful symbol. It tells you to multiply by every number that is above it and divide by every number that is below it. And it’s as easy as that. So when you are simplifying expressions like the ones in the two examples above or the more complicated example at the beginning of this post, all you need to do is to put numbers that are multiplied above the vinculum and numbers that are divided below the vinculum. And what happens if you are multiplying or diving by a fraction? Then put the numerator on top and the denominator on the bottom if you are multiplying. Flip the fraction over first if you’re dividing (as I did in Example 2). When you do this, you will end up with a 2-stack fraction that can be easily evaluated. Let’s look at the original problem again:

\dfrac{(2)(3)}{4}=\dfrac{(5)(6)(7)B}{(8)(9)}

To solve for B, you will multiply by 8 and 9, so they go on top. And you will divide by 5, 6 and 7, so they go on the bottom. In one step, you’ve solved for B as follows:

B= \dfrac{2 \cdot 3 \cdot 8 \cdot 9}{5 \cdot 6 \cdot 7}

Compare that to the monstrosity at the beginning. So much easier!

P.S. You don’t really have to remember the name “vinculum”. Most people don’t know what the line is called and don’t care. Chances are good your math teacher doesn’t even know the term. To make things even more bizarre, when you write a fraction with a slash instead of a horizontal line like this—2/3 – the slash is called a “virgule”.  Most people don’t know that one and don’t care about it either. All you have to remember is that the horizontal line in a fraction tells you to multiply on top and divide on the bottom.

Algebra 1, Algebra 2, Calculus, Physics, Precalc/Trig
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