Last week, in Fundamental Understanding of Mathematics LII, we fleshed out a balance model for symbolic manipulation of equations by adding a rule for modifying only one side of an equation by adding a zero pair to that side.
This week, I thought I'd put the model to use by solving some problems with it.
- Irwin walks down the sidewalk with three cents in his pocket. He sees a nickle on the sidewalk and picks it up. How much money does Irwin have now?
Ok, this is straightforward. We want to know how much money Irwin has, so we'll make that amount the banana. The problem tells us that the banana is equal to three cents Irwin had to begin with, plus five cents (the nickel) he picked up.
We simply add up the numbers on the right hand side, and discover that Irwin has eight cents.
- Adrienne is writing a haiku. So far, she has
Trying to learn math
Reading Orinoco words
banana
How many syllables must she add to the last line?
To solve this one, we need some outside knowledge: the form of a haiku -- five syllables, seven syllables, then five syllables, and how many syllables in ba / na / na (three.)
Let's use the apple for the unknown number of syllables needed for the last line. That number, added to the syllables already there, must equal five.
We want to get the apple by itself, with the known numbers on the other side. We need to get the 3 off the left hand side, we can do that by subtracting 3 from both sides.
The right side is straightforward arithmetic: 5 - 3 = 2. The left side is a bit trickier.
First of all, since we don't know what apple plus three is, we use the associative property for addition [ (a+b)+c=a+(b+c) ] to group the 3 - 3 and do that subtraction first. 3 - 3 = 0
Then, we use the additive identity (a + 0 = a) to replace apple plus zero with just apple
to get to our answer: Adrienne needs two more syllables, either two one-syllable words, or one two-syllable word.
- Ralph can buy two pencils. If he had another nickel, he could buy three pencils. How much is a pencil?
Another money problem. Since we want to know the price of a pencil, we'll use a banana to represent that amount of money. Ralph can buy two of them, so he has two times that amount, or two bananas worth.
If Ralph had another nickel, then he'd have more money, so we'll add that hypothetical nickel to the money he already has
And, the problem tells us, if Ralph had that other nickel, he'd also be able to buy three pencils, so we can make the money he has (real and hypothetical) equal to three pencils worth of cash, or three bananas.
Now the scales are balanced, and we can start using our equality rules to figure this one out.
We want unknowns on one side, known numbers on the other, we can get there by removing two bananas from each side.
And we're done. A banana is worth a nickel. I mean, a pencil is worth a nickel.
- Joey is three times as old as his sister Megan. In five years he will only be twice as old. How old is Megan now?
We want to know Megan's age now. So, let's assign an apple as an unknown number representing Megan's present age. The first part of the problem statement says Joey's age is three times that apple.
But what goes on the other side? Well, we don't know the value of three apples, so we have another unknown number -- Joey's age. Let's use a banana to represent that number.
Here is our set up for "Joey is three times as old as his sister Megan." There are no known numbers to do calculations with, and the unknowns are on opposite sides, so, if we need to, we can use this statement to substitute three apples for a banana, or vice versa. There's little else we can do with this now, so let's move on to the rest of the problem.
In five years Joey will be twice as old as Megan.
In five years, Megan will be apple plus five years old,
and Joey will be banana plus five years old.
Two times Megan's future age is equal to Joey's future age
Here's our second set up for this problem. This we can do some things with. To begin, let's use the distributive property of multiplication over addition [ a(b+c) = ab + ac ] to get rid of the parentheses on the left side.
Now there are several ways we could proceed. We could work on getting the unknown numbers on one side, and the known numbers on the other. Or, recalling that we can substitute three apples for one banana, we could work to get either one banana or three apples alone on one side. Or, we could get rid of the banana straight off, by substituting three apples for it, and be left just working with one unknown number.
Which is the "right" way?
As long as we use number properties correctly, use only information given in the problem, and keep balanced equations balanced, any of these ways is the "right" way, because they will all lead to a correct answer.
So, I will show the first method, and leave the other paths for the readers.
Get the unknowns on one side, and the knowns on the other. I will subtract five from both sides, which will leave a known number on the left side, so I will also subtract two apples from both sides, which will remove them from the left side, leaving knowns on the left, and unknowns on the right.
The left side is easy arithmetic, 10 - 5 = 5. The right side becomes easy when I recall that a banana is worth three apples, and three apples minus two apples is just one apple.
So our apple is five, since that represents Megan's age now, Megan is five years old. Joey then would be fifteen, and in five years Megan will be 10 and Joey will be 20. This meets the problem criteria, so our answer checks out: Megan is five years old.
Have fun in the comments.