Let’s go on a road trip!

When I teach geometry, especially geometry involving angle measures like this problem, I like to describe the process of solving a problem as taking a little road trip. I describe it this way because this is how I personally feel when solving a problem like this – my eyes rove around the figure from one intersection to the next, and I hop in my little math car and drive along lines and stop at intersections to figure out where I am. Geometry is a very visual discipline, and as a visual learner, I have the most fun when I can trace a physical journey around the problem, solving things as I go. So let’s hop in our math car and chase this problem down!

The first step in any problem like this is to sort out what our givens mean. This usually involves a bit of deciphering the image as well; making sure it unpacks and makes sense in your head before you progress further. You have to be able to read the map if you don’t want to get lost in the middle of Illinois with nothing but cornfields for miles! So, in this figure we’ve got two parallel lines, l and m, that are cut by two parallel transversals, n and o. The figure in the center of the lines is a parallelogram – which may or may not be relevant to the problem, but it’s a good thing to state at the outset. My approach is always to state everything you can deduce from the givens as soon as the givens appear; you can always ignore the information later if it’s not needed.

We’ve also been given the values of two of our angles, angle 1 and angle 13, but they’re written as algebraic expressions rather than straight-up numbers. This is where I always ask the student, “Why would they give us the information like this? Why not just tell us the numbers? What are they expecting us to deduce from this?” If the student can answer those questions, they’re halfway through a proof already. In this case, the angles given are a pair of corresponding angles – the upper-left corner of two of the intersections. What do we know about corresponding angles? They’re congruent, which means their measures are equal. So we should be able to simply set those two expressions equal to each other and solve for x. That’s a good place to start, so let’s hop out of the car and do that.

10x – 7 = 7x + 29
10x = 7x + 36
3x = 36
x = 12

Okay, so x is 12. That’s progress, right?

Right?

Well, how does that help us figure out angle 12? Is there some way we should be able to figure out angle 12 just by knowing some of the other angles? Let’s hop back in the car and head over that way to look closer.

Well, looking just at line o, we see another set of corresponding angles – angle 13 should be congruent to angle 9. That’s good, we’re at least at the correct intersection now. Now to get our bearings. Given a single intersection, we basically have two angle measures repeated twice – you can think of them as the pairs of vertical angles, or the pairs of supplementary angles, whichever makes more sense to you. Either way, angle 9 should be congruent to angle 11, and both 9 and 11 should be supplementary to angle 12 and angle 10. That means, whatever value angle 9 is, we can subtract that from 180 to get the value of angle 12.

Okay, sweet. We have a plan. Let’s hop back in the car and go figure out what angle 13 is so we can bring that with us to angle 9.

*road trip montage*

Well, we know what x is now, right? So let’s just plug that back in to our given and see what we get.

X = 12
angle 13 = 7x + 29
7(12) + 29
84 + 29
113

Okay, so angles 1, 13, and 9 should all be 113. (Incidentally, so should angle 5, but that’s for another problem.) Let’s throw 113 in the trunk and drive back over to angle 12!

*changes the radio station*

Let’s call angle 12 ‘y’ for clarity. Angle 12 is supplementary to angle 9, so the two should add up to 180.

180 = 113 + y
y = 180 – 113
y = 67

So angle 12 should be 67 degrees. Problem solved!

Now, for a bit of extra credit…

These types of problems often involve one image that has several different problems to go along with it. If you had other problems to solve with this same figure, you’d be thrilled to know that you now know the values of each and every angle in the figure. They’re all either 113 or 67, set up in supplementary and/or vertical pairs with the existing ones. Finish your road trip and jot down the values of all the other angles in case you need them later!

Angle 12 is 67, which makes angle 10 also 67 and angle 11 113.
*road trip montage*
Corresponding angles makes angles 5 and 7 113, and therefore angles 6 and 8 are 67.
*pit stop for gas and snacks, road trip continues*
And so on until your road trip is finished and your figure is entirely labeled. That’s the power of parallel lines!

Sometimes the same procedure shows up in two different contexts. This is especially common in the fields of math and science, as science employs in real-world application many of the techniques we learn in their abstract form in math class. For some reason, the principle as shown in a high-school science class is often much harder for students to understand than it was in the math class. (My personal theory is that science teachers are applying the concept in a way that changes how they explain how it works, and they probably have not collaborated with the student’s math teacher to ensure they’re reinforcing the same terminology.) Last week one of my students ran into this phenomenon in her own work; a concept from last year’s math class showed up in her physics class. To help her understand it, we went back to the original math concept and talked about proportions.

The science homework she was struggling with was the old chestnut about unit conversions; rows and rows of fractions set up in perplexing fashion, with equals signs between each one and confusing scribbles across and between them. When taught in physics, the idea of unit conversions frequently becomes clouded and nonsensical – it becomes a feat of rote memorization to attempt to figure out what the correct procedure is for each specific case. In truth, unit conversions are a simple case of mathematical proportions, and the easiest way for my student to understand it was to connect it to the original context in which she learned about proportions last year – similar triangles.

A proportion is a pair of fractions that are set equal to each other. Remember – one of the definitions of a fraction is a ratio (comparison of two quantities), so setting two fractions equal to each other signifies that these two ratios are equivalent. We use this to understand similar triangles in early algebra, working through the logical reasoning process to grasp the concept that increasing all parts of the figure by the same amount will maintain the relationship of parts to each other. Unit conversions are simply another application of this same idea.

Let’s say your physics problem involves the length of a bridge. The problem states that the bridge is 3200 feet long, and wants to know how many meters long that is. We’re not doing anything to change the bridge itself, we’re simply changing how we measure it. That can be analogous to finding an equivalent fraction – we’re not changing the value of the fraction, just how it’s written. No matter whether we measure the bridge in feet, meters, inches, or miles, the relationship of the various parts of the bridge to each other (the proportions of the bridge) will stay the same. So this is a proportion problem.

For a proportion problem to work, we need two fractions. We only have one number in the problem, so where are the rest of the puzzle pieces? They’re implied, in the form of something called a conversion ratio. The conversion ratio is the fraction which indicates how the two units relate to each other – in our case, how many feet are in one meter. If we know that, it’s a simple similar-triangles-type problem to figure out our conversion.

So what’s the conversion ratio for feet into meters? Well, usually in a problem like this, the conversion ratios are either easy to find out for yourself, or they’re provided for you. A quick internet search reveals that one meter is equal to 3.28084 feet. So, without doing any math just yet, common sense tells us that converting the 3200 feet of the bridge into meters should give us a smaller number than we started with – there are multiple feet in one meter, so we’ll be dividing the number of feet up into one-meter chunks. (I find this kind of common sense flagging is helpful for these kinds of problems.)

So let’s figure out our proportion!

1 meter/3.28084 feet = x meters/3200 feet

I find it helpful to read this sentence to yourself as an analogy: “One meter is to 3.28084 feet as x meters are to 3200 feet.”

We know we have the proportion set up correctly because the matching units are on corresponding sides of each fraction. Just as a similar triangle problem needs the corresponding sides to match up with either the top or bottom of each fraction, unit conversions need to match up the units the same way.

You know what to do from here, right? Cross-multiply!

3.28084x = 3200(1)

x = 3200/3.28084

Hold up there for just a second. Remember a bit further up when I flagged for us that we’d be “ dividing the number of feet up into one-meter chunks”? Check it out – that’s exactly what we’re doing! Taking the total number of feet and dividing it by the number of feet per meter. That’s another way it’s sometimes described in science classes:

total feet [divided by] feet/meter

and when you divide by a fraction, you multiply by the reciprocal, so:

total feet * meter/feet

Sometimes science teachers will make a big show out of placing the fractions like this, so that they can physically cancel out the feet and all they have left is meters. I’ve found this to be helpful in calculations that require multiple conversions (inches to kilometers, for example), since it makes sure you keep the units straight, but it’s just as easy to do the proportion problems one at a time, and these days you can usually find the conversion ratio to go straight from one to the other anyway. A quick internet search tells me the conversion ratio for the example I just gave is 39370.1 inches per kilometer. Boom. Done. Write your proportion and go to town.

Anyway, heading back to our original problem, we see that:

x = 3200 feet/3.28084 feet per meter

x = 975.36 meters

Which does satisfy our initial common sense check, as it’s considerably smaller than the starting value.

In these examples, I find it much more useful to think of them as proportion problems in the vein of the similar triangle problems we practice so much in algebra. The algorithm is the same, it’s just the context that differs, and if that’s the case, why not think of the problem using whatever context makes the most sense to you?

To learn more about similar triangles, check out my blog post about that time my TV broke.

A few summers ago I wrote a blog post about finding math in unexpected places as a way to keep skills sharp through the summer break. One of the unexpected places I talked about was the world of tabletop Role-Playing Games (RPGs) such as Dungeons & Dragons. Such games are essentially communal storytelling exercises which use chance elements to help guide the story via a set of polyhedral dice.

I’ve been running a D&D game for a group of friends for several months now, serving as “Game Master.” As Game Master I serve as lead storyteller for the group, while the others each create a character to experience the story firsthand. My job is to create the framework for the story. I devise and flesh out the world that the story takes place in, present challenges for the players to overcome, and rationalize the effect their actions have upon the world. Overall, my goal is to create circumstances that will allow the players to be heroes. Today I’d like to delve a little deeper into the math applications involved in a D&D game, through the use of an example from the game I’m currently running.

In the current story arc, my players recently made the acquaintance of a tribe of goblins led by a goblin chief by the name of Skizaan. Goblins are a very classic antagonist in the setting of D&D, and they are often seen as little more than cannon fodder – weak enemies who can be bowled over easily. However, these particular goblins are highly organized and skilled at group tactics, and as the brains behind their operation I wanted Skizaan to present more of a challenge to the group. In the rulebook for the game there are statistics for the average goblin, as well as a “Goblin Boss” – both of which were a bit too weak for my tastes. I decided to amp up Skizaan by altering some of his statistics to increase his ‘Challenge Rating’ (known as ‘CR’ – the system the game uses to help Game Masters determine which antagonists will present an appropriate challenge for a player group of a given power level).

As part of this alteration, I needed to increase the amount of damage Skizaan dealt with weapon attacks to bring it in line with my target CR. The rulebook contains a handy table listing the average values for various statistics at different CRs, so looking at the table entry for my target CR shows me that the average damage value per round should be about 33. Great – that’s my target. However, there’s a bit of a wrinkle – D&D uses dice rolls to determine damage dealt, so every enemy lists a number and type of dice to roll when an attack hits. This adds another element of chance, where combat can lead to drastically different outcomes depending on the luck of the dice – which is one of my favorite aspects of the game. The table only lists average damage output, though, not dice ranges, so I needed to figure out what combination of dice would result in an average damage of 33.

This is starting to sound like a word problem from a math test!

Now, there are several ways to do that, but here’s the system I worked out. To figure out the average result of a die roll, I just need to take the average of the lowest possible roll and the highest possible roll. Since I know my target average, reverse engineering the average formula is the quickest way to get there algebraically:

(lowest (l) + highest (h))/2 = average

(l + h)/2 = 33

l + h = 66

Now it’s time to think about the way dice rolls work for a moment. The lowest possible outcome would be all 1’s on the dice, the highest would be every die rolling its maximum. The game uses dice that have 4 sides, 6 sides, 8 sides, 10 sides, 12 sides, and 20 sides, so I have a few options. The thing that immediately hits me, though, is that 66 is 11 times 6. If I went with six dice, my lowest possible roll would be 6, regardless of what kind of dice I was using. That would leave me with 60 points left to divvy up between 6 dice, so they’d need to be 10-sided dice. In game terms, I’d write that as 6d10. So rather than the original 1d6 of the Goblin Boss, my new amped-up goblin chief could do 6d10.

So that’s one option, but there’s another aspect of the luck of the dice that I want to acknowledge: the spread of possible outcomes. The lowest and highest values of that 6d10 roll would be 6 and 60, which means that this attack could vary wildly in terms of how much damage it actually does. Sure, on average it’ll be around 33, but if my dice are having an off day when we actually play, Skizaan could do practically no damage and the players would steamroller him, which would be pretty anticlimactic. I’d like to find a way to make the rolls more consistent, with less variation. The way to do that is to change up the type of die rolled to decrease the spread – make the highest and lowest values closer together.

To do that, I’ll need to shift the composition of the roll to be more dice with fewer sides each. More dice increases the lowest value, and less sides each decreases the highest value. So let’s take our target of l + h = 66 again. We can split it into the same 6 and 10, but switch their placement – instead of six ten-sided dice, let’s use ten six-sided dice. That would make the lowest value 10 and the highest value 60. That would increase the average value a bit, from 33 to 35, but would condense the spread a bit in the process.

However, there might be an even better option – we do have four-sided dice at our disposal. Let’s say we used as many four-sided dice as possible to get the same average outcome. At this point, I use a slightly different logical process. Take a look at our previous two examples. With ten-sided dice, we ended up with six times eleven. With six-sided dice, we ended up with ten times seven. In each case, we end up with the number of dice rolled multiplied by one more than the number of sides on the dice. So if we have four-sided dice, we can figure out the number of dice to roll by dividing our target of 66 by one more than the number of sides on the dice – 66 divided by 5. Rounding down to avoid decimals, we end up with 13 four-sided dice. Plugging that back into our main formula to check our work, we get

(13 + 52)/2

32.5

So we’ve decreased the average just a tiny bit, but in exchange we’ve given ourselves a much more condensed spread on the dice. Minimum damage for Skizaan’s attacks is now 13, which is much better than 6, and his maximum damage potential is a still very threatening 52.

This is just one example of the many ways in which serving as Game Master for a game of D&D involves quite a bit of math and logic. Running this game has been a great experience for me in problem solving with algebra, and I thought you might enjoy hearing about it.

We’re going back to basics today with a Math Journey covering the three broad categories of symbols. I’ve found this concept very handy when introducing Algebra to middle school students. So let’s go!

Math is a language, and I find it often helps to think of it as such right from the beginning. Just as there are different parts of speech in a language, so there are different ‘parts of speech’ in math. Where a spoken language includes parts of speech such as nouns, verbs, and adjectives, math has three major types of symbols: constants, operators, and variables. Let’s go over each one in detail.

Constants
These would be the equivalent of your nouns. A Constant is a number – it has a single, discrete place on the number line. Even if the number itself is ugly – a non-terminating decimal, for example – it still does exist in a specific spot somewhere on the number line. In addition to the obvious constants, math frequently uses what I refer to as ‘special constants’ or ‘named constants’ – ugly numbers that are important enough for some reason or other that mathematicians have given them special names and symbols. Pi is a good example of this; mathematicians figured out that performing a specific calculation on a circle always yields the same number, regardless of which circle is used, and figured that that number was special enough to warrant a name. In much the same way, other constants such as e and i have been given names and special symbols to represent them due to their importance for certain calculations. But the important thing to remember here is that all of these named constants do have specific spots on the number line – they don’t change value depending on the situation. Pi will always be approximately 3.14159, no matter what you do to the rest of the problem.

But that’s not always the case.

Variables
Variables also represent constants, however in this case the actual value of the constant is unknown. The variable does have a specific spot on the number line, but we don’t know where it is. Its location on the number line can vary from problem to problem, but within a single problem it is always consistent. We generally refer to variables using lowercase letters, traditionally starting with x, y, and z, and then moving to others if necessary. In practice, a variable behaves just like a constant, since it does actually represent a constant. It can be manipulated the same way you would a constant, except of course you don’t know the value so you’ll have to leave some calculations unfinished until you get to a point where you can identify the mystery number. Funnily enough, many elementary math programs use the concept of variables, but they don’t define them as such. If you’ve ever seen a basic math worksheet with a question mark in a problem, you’ve seen a variable. All algebra does is change over from using a question mark to a series of lowercase letters.

3 + ? = 7
3 + x = 7

Operators
All the constants and variables in the world won’t help us without an operator. Remember how your grammar teacher was always going on about how every sentence needs a verb? Well, every mathematical sentence or phrase needs an operator. An operator is a symbol that performs an action on a constant or set of constants. Plus signs, minus signs, multiplication and division symbols are all operators, but so are square root bars and fraction bars. In fact, as you may have read in one of my earlier Math Journeys, a fraction is just an indication of the top constant being divided by the bottom constant. The equals sign is also an operator of sorts, though it doesn’t perform an action on the constants so much as declare a relationship between them. The greek letter Sigma is an operator as well, used to represent taking the sum of a series.

And then there’s the special operator known as ‘a function.’ We’ve talked about functions multiple times before in my blog, and I usually introduce it as a machine that turns one number into another by applying a set rule. That sounds like an operator to me! The key here is that a function is kind of a general operator, one that you can define within a given problem any way you want or need it to be. Want to indicate a specific sequence of operations performed on a number repeatedly over the course of a single problem? Use a function and define it appropriately!

Breaking down the world of math symbols in this way helps to clear up some of the confusion that often results from the particulars of traditional naming conventions. Consider, for instance, the following six symbols:

e ∏
x Ѳ
f() ∑

All three in the first column are lowercase letters, and all three in the second column are greek symbols. However, their usage in math is better represented by the horizontal rows. The first row are constants, the second variables, and the third operators. And the way they behave differs accordingly. So the next time you’re confused, take a look at which type of symbol you’re working with!

Standardized test math doesn’t behave like normal math. On a normal math test, your knowledge of the concepts and material is being tested, using (hopefully) fair test questions. On a standardized test, though, they’re looking for you to think outside the box, to apply math concepts and algorithms to unusual situations, and to really understand what they’re looking for and find the quickest way to go about it. Let’s take a question from a recent GRE student’s lesson:

If 4x – 5y = 10 and 6y – 3x = 22, then what is x + y?

Now, this is a set of two equations with two variables each, so it looks to me like a perfect candidate for solving as a system. If I were solving this one on a regular math test, I’d start off trying the substitution method, since I’m more comfortable with that one. So let’s explore that one first:

I’ll start by solving the first equation for y:

4x – 5y = 10
– 5y = 10 – 4x
y = (-10/5) – (4/-5)x
y = -2 + (4/5)x

Then I’ll plug that in for y in the second equation:

6(-2 + [4/5]x) – 3x = 22
-12 + (24/5)x – 3x = 22 Now we have to convert the 3x into a fraction
-12 + (24/5)x – (15/5)x = 22
-12 + (9/5)x = 22
(9/5)x = 34
x = 34 (5/9)
x = 170/9

Then plug that back in for x in the first equation:

y = -2 + (4/5)(170/9)
y = 136/9

And, FINALLY, find the quantity asked for in the problem by adding x and y together:

x + y = (170/9) + (136/9)
x + y = 306/9
x + y = 34

Well, that’s one way to find the answer, but that took a long time, with lots of large numbers, and lots of potential for mistakes. This is a standardized test, remember, so time is a factor here. Take a look at the question again. It’s asking for x + y. Why wouldn’t it be asking simply for x, or y, or even x and y, for that matter? Is it because x + y is a much cleaner number? Is it to be ornery? To make you waste time?

Well, to be honest, the answer to that last question is yes, but not in the way you might think. In our math classes, we’re hardwired to try to solve for x – we want to end up with a nice clean number to equal one of our variables. It’s the way most math classes work; manipulate the equation until it tells you the missing piece of information. The test builders know that, and they know that everyone’s first instinct in a math problem is to try to solve for x. But in this case, they’re not asking for the value of x; they’re asking for the value of an expression containing x. And they’re doing that very deliberately – because finding x + y is much easier than finding x.

Take a look at our system again – this time I’ll re-arrange it slightly in preparation for using the addition method to solve it:

4x – 5y = 10
– 3x + 6y = 22

See it yet? Use the addition method – don’t even modify anything – and add straight down the columns:

4x – 5y = 10
– 3x + 6y = 22

–> x + y = 32

Well, would you look at that? That’s the answer they’re looking for – and you’ll notice it’s not the same answer as our previous attempt. Not only would you have wasted a bunch of time going through all those hoops to solve with substitution, but you would have gotten the question wrong to boot!

It’s an odd way of looking at a math problem, but one of the biggest strategies I tell my students is to not think about the test as a math test. It’s a logic test that happens to involve numbers. Here, the test is remembering to only solve for what you need. Don’t bother getting all the way down to x if x won’t help you in the end. Sometimes they’re asking for a quantity because going any further past that quantity will only cause you grief. Solve for the quantity they ask for, and no more.

I received this problem from a friend, who was having trouble while helping her nephew with it. It turned out to be quite a doozy, so I’m presenting it as today’s Math Journey to show how the process we used last time works even with a gnarly, complicated problem.

Solve using the Addition Method:

3x – 3y + 4z = – 15
3x + y – 3z = – 8
23x – y – 4z = 0

As we discussed last month, the basic idea behind solving a system of equations is to use one equation to solve another for a specific variable, and to do that enough times that you can eventually rewrite one of those equations with only one variable in it, and solve from there. The way I learned to do this is the “substitution” method, where you solve one equation for one variable, plug the expression in for that variable in a second equation, et cetera until you’re down to one variable. The addition method works a little differently, but it’s the same basic goal: eliminate enough of the variables that you can solve the system. The basic idea of the addition method is to figure out a way to get two of the equations to have opposite coefficients on one of the variables, so that if you were to add all the terms in those two equations, you’d cancel out one of the variables entirely.

Take a look at the first and third equations for now:

3x -3y +4z = -15
23x -y -4z = 0

The z’s have a positive 4 and a negative 4 as coefficients, so if we were to add the third equation to the first equation, which in practice means adding straight down each column in the equations, we’d get:

26x -4y + 0z = -15

So the z’s will cancel each other out. Now, our new first equation is:

26x -4y = -15

Good; we’ve eliminated one variable from the first equation. Go back to the original system. Look at the second and third equations. Do you see any way to eliminate the z’s using just those two equations in the system? I do. It’s not as immediately apparent, but it’s there.

3x +y -3z = -8
23x -y -4z = 0

This one we have to do some preliminary work for. If we’re going to add them, we need one equation to have a positive coefficient and the other a negative, and they need to have the same absolute value. They’re both negative right now, and the numbers are different. No problem – think about finding a Lowest Common Denominator back in middle school. Remember, we can multiply an entire equation by a single constant without changing the value of the equation, just like finding an equivalent fraction. We’ll have to do both cases for this one – sometimes you can get away with one, but not today.

Think LCD – what numbers are we going to multiply each equation by to eliminate the z’s?

The first one should get a 4, and the second one a 3, and one of them needs to be negative. Doesn’t matter which one, so let’s pick. I say negative 3 and positive 4.

4 (3x +y -3z = -8) → 12x +4y -12z = -32
-3 (23x -y -4z = 0) → -69x +3y+12z = 0

So now, when we add straight down, we get:

-57x +7y +0z = -32

Or:

-57x +7y = -32

Okay, now let’s go back to our original system. Which equation did we use in both of our additions? The third, right? Okay, so set that one aside for now. We’re going to replace the first and second equations with our new ones:

26x -4y = -15
-57x +7y = -32

Now, we can treat these two as a system of two variables in two equations, and repeat the above process to reduce it down to one variable. I’d say, since the coefficients are smaller and easier to handle, let’s try to eliminate y. Same process as before – what’s our LCD? 28. So:

7 (26x -4y = -15) → 182x -28y = -105
4 (-57x +7y = -32) → -228x +28y =-128

When we add those straight down, we’ll get:

-46x = -233

And THAT we can solve for x, since it’s only got one variable:

x = 233/46 (I’ll leave it as a fraction, since it’s ugly)

Now, we can do what’s sometimes called “back-solving” to get our final answer.

First, plug that into the first of our new (2-variable) equations to solve for y:

26x -4y = -15
26(233/46) -4y = -15
(3029/23) -4y = -15
-4y = -15 -(3029/23)
-4y = -(3374/23)
y = -(3374/23) / -4
y = 1687/46

Now that we have y, we substitute BOTH our x and our y into the third equation (the one we set aside at the beginning – the one with three variables still in it) to solve for z.

23x -y -4z = 0
23(233/46) – (1687/46) -4z = 0
(5359/46) – (1687/46) = 4z
3672/46 = 4z
z = (3672/46) * (¼)
z = 918/46

or, when you reduce it all the way: 459/23

So, our final values for all three variables are:

x = 233/46
y = 1687/46
z = 459/23

And yes, they’re ugly, but yes, that’s it.

Just to be sure, I solved this same problem using the substitution method, and got the same answers. It’s a bit more convoluted to use the addition method, but it still works. (My guess is that’s why the problem specifically required you to use the addition method – it would have been too easy the other way!) As with Lowest Common Denominator problems, if you don’t have compatible numbers given to you the fractions are going to get big and ugly really fast. This one was unfortunate in that 23 is prime, so all the numbers were forced into being pretty big. Don’t worry though; you’ll still get an answer eventually.

Incidentally, systems can be tricky to get the hang of at first because any given system has many different paths you could take to get to the end. You can start with whichever equation you want and solve for whichever variable you want. Ideally, the problem should work no matter what order you choose to use the equations in. Some paths are shorter or easier than others, though, so it pays to think briefly about how to tackle a new problem to save yourself a headache later on.

Settle in, folks, today’s a long one.

In The Function Machine, we learned that functions can be depicted as curves graphed on a coordinate plane. In What Does the Function Look Like?, we learned how to tell the general shape of a function’s graph based on characteristics of its equation, and vice versa. Today, we’ll be focusing on linear equations (meaning any equation that graphs into a straight line).

The defining characteristic of a linear equation is that the highest power of x in the equation is x to the first. This denotes that for every y value, there is exactly one corresponding x value. Of course, there is always exactly one corresponding y value for every x, but this is one of those “square is a rectangle; rectangle is not necessarily a square” moments. We know there’s exactly one y for every x because we choose our x’s independently and the y’s are dependent on them. There can’t be more than one y for any given x; you’ve only got one output slot in your machine. We don’t necessarily know that our y values won’t repeat, though, unless the function only has x to the first power and no higher.

So let’s look at a couple of linear functions. I’ll pick two at random.

y = 2x + 3
y = 6x – 5

Now, let’s think for a moment about how those two lines might relate to each other. There’s only a couple of ways a pair of lines can interact. They could be parallel, meaning that they never intersect. Parallel lines would have the same slope (you can find the slope as the coefficient on the x value in the equations above). These two lines do not have the same slope, so they’re not parallel.

If they’re not parallel, then by definition, they must intersect. And since they’re straight lines, they intersect at exactly one point. Let’s think for a moment about how we are able to graph equations in the first place. For a point to be on the line of an equation, that means that when you choose that point’s x value as your input, you get that point’s y value as your output. The x and y values of that point, plugged in to the equation, would make it true.

Well, if the two equations above intersect, then that means there’s exactly one point where the same set of x and y values would make both of these equations true at the same time. That sounds like something that’d be handy to know, huh? Wonder how we could do it…

What we’ve got here is known as a system of equations. That just means a set of more than one equation that we’re looking at together. And finding that intersect point requires what we refer to as “solving the system.” There are a couple of ways to do it – we’ll just look at the two most common.

Method 1: Substitution Gone Wild

We’ve already established that if you have a given value for one of the variables, you can plug that value in for that variable in an equation to find the remaining one. Extrapolate that – if you can substitute individual values for a variable, why couldn’t you substitute whole expressions in for variables, if you happened to know an appropriate one? Well, turns out you can. Let’s look at our equations again.

y = 2x + 3
y = 6x – 5

Now, the first step to substitution is to solve one of these equations for one variable. They’re both already solved for y, but for the sake of demonstration, I’m going to solve the first one for x.

y = 2x + 3
y – 3 = 2x
½ y – 3/2 = x

So now we have x in terms of y. Let’s take that newly-found expression and substitute it in for x in the other equation.

y = 6x – 5
y = 6(½ y – 3/2) – 5

Now we have an equation with only one variable in it. THAT we can solve to get a number. So let’s do it.

y = 6(½ y – 3/2) – 5
y = 6/2 y – 18/2 – 5
y = 3y – 9 – 5
y = 3y – 14
14 + y = 3y
14 = 2y
7 = y

So y is 7. Now that we have that information, we can plug THAT back in to our first equation to find x. The trick with this method is very simple – ALWAYS SWITCH EQUATIONS! If you accidentally try to plug this number back into the equation you were just using, you’ll end up in an infinite loop, going around in circles and getting more and more frustrated. That or you’ll cancel all the variables out and end up with 7 = 7, which doesn’t help at all. Remember to keep switching back and forth. Back to the first equation we go!

y = 2x + 3
7 = 2x + 3
4 = 2x
2 = x

So the intersect point of these two equations is at (2, 7). That’s the value that makes both equations true at the same time.

I like this method for its simplicity and ease of use. But there is another method that’s pretty popular, and it all hinges on opposing coefficients.

Method 2: Addition/Elimination/Subtraction method – Opposing Coefficients

This method goes by many names. I prefer the Elimination method, because it tells you what you’re doing. The goal with this method is the same as the previous one – get one equation to only have one variable in it. But this time, we’re going about it a little differently.

The basic idea of the Elimination method is to figure out a way to get the two equations to have opposite coefficients on one of the variables, so that if you were to add all the terms in those two equations, you’d cancel out one of the variables entirely. It requires us to establish at the outset that a couple of math moves are legal:

1. That we can multiply an entire equation by a single constant, multiplying each term by that constant, and we won’t change the value of the equation itself. It’s just like creating equivalent fractions back in middle school; as long as you do the same thing to each term, you’re creating an equivalent equation that looks different so that we can work with it better.
2. That we can add two equations together, by adding corresponding terms, to create a new equation that will also be true. In practice we’ll be adding straight down in columns.

Let’s take a look at our equations again.

y = 2x + 3
y = 6x – 5

Okay, I see some potential here. If we were to multiply the first equation by -3, we’d end up with a -6 on the x that would cancel out with the positive 6 in the second equation. Let’s try it.

-3(y = 2x + 3) → -3y = -6x – 9, so

-3y = -6x – 9
+ y = 6x – 5
-2y = 0x – 14

So we now have zero x’s – they’ve cancelled themselves out. Let’s solve the remaining equation for y to see what we get.

-2y = 0x – 14
-2y = -14
y = 7

Same as before. That’s a good sign! Now, here’s the only tricky part. We have to keep switching equations, so we need to know which one we were technically using up there. Which equation were we actually using? Well, the way I think of it is this: we added the second to the first, but we didn’t DO anything to the second equation, so that one’s still fair game. Think of it as using the second equation as a key to unlock the first one. We haven’t actually worked with the key equation yet, just used it to unlock others. Let’s plug our new value for y back in to the key equation.

y = 6x – 5
7 = 6x – 5
12 = 6x
2 = x

So our point is (2, 7) – just like with the other method!

And if we put those two original equations into a graphing calculator, we get:

Thanks to Desmos Graphing Calculator for the image.

Well, fancy that!

To be honest, I’d rather use the substitution method. I think the process is simpler to remember, and with systems of equations, there are so many different ways to get from start to finish that I’d rather have a simple process I can remember than try to keep straight which equation is “active.” Plus, if Calculus is in your future it helps to learn the basic idea of substituting an expression in for a variable now, while it’s still relatively simple, so that later when you’re writing derivatives that stretch all the way across your notebook page you’ll have a more instinctive sense of how to accomplish the substitution part of it.

Elimination also has a tendency to turn into big ugly fractions – notice I chose coefficients for the terms that were both small and factorable? This is a nice, simple example, but math textbooks have a tendency to make problems unnecessarily complicated simply by using ugly numbers.

Speaking of which, stay tuned for next month’s Math Journey, when we tackle a big ugly system in three variables!

Since it’s Thanksgiving week, let’s think about pie for a second. No, not mathematical pi, just actual real edible pies. For Thanksgiving I’m in charge of making dessert, so I’ll be bringing two pies, one pumpkin and one apple. Let’s say that I sliced the apple pie into 12 pieces, and the pumpkin pie, since it held together better, into 18.

Fast forward to the end of the evening. My pies were a big hit, and I have almost none left. In fact, all I have is three pieces of apple and four pieces of pumpkin. I want to combine the remaining slices into a single pie pan, so that they take up less space in the fridge. How do I figure out if my remaining pie will fit in one pan?

Well, let’s start by writing down the remaining amounts of pie in the form of fractions. Remember, one of the definitions of a fraction is parts of a whole, so let’s apply that definition to figure out our starting fractions.

The apple pie was cut into 12 pieces, and we have three out of twelve left. So our apple pie fraction is

 3
12

The pumpkin was cut into 18 pieces, and we have four left. So the pumpkin fraction is

 4
18

To figure out how much pie we have left and whether it will fit into one pan, we’ll have to add these two fractions together.

 3  +  4
12    18

Now, the first problem we run into is that our pies were cut into different amounts of pieces. We can’t accurately compare the two pies until we subdivide them into the same size pieces. So to do this, we’ll use a process that often confuses students when they first learn it: finding the Lowest Common Denominator.

I think this concept confuses students mostly because the terms used are long and complicated. So let’s talk about Lowest Common Denominator for a moment. We only specify Lowest because there are many different denominators that would work for this problem, but most of them are bigger than they need to be. We don’t want to work with numbers that are any larger than they need to be, so this process ensures that we don’t have to do any extra reducing after the fact.

We need to find a number that can divide evenly by both of our current denominators (though not necessarily at the same time). So all of the factors involved in either of our current denominators need to be present in our LCD. Let’s take the foolproof, long-way-round method and break each denominator down into its prime factors. You can do this with a factor tree:

12                18
3 x 4            3 x 6
3 x 2 x 2      3 x 3 x 2

So we see a lot of overlap here between the two denominators, right? Both contain a 3 and a 2. This is where the “lowest” part of the Lowest Common Denominator comes in. Worst case scenario, we could simply multiply each denominator by the other one and we’d come up with a number that both would divide by, right? But look at that overlap. That denominator might be common, but it certainly wouldn’t be the lowest – you’d be able to divide both top and bottom of each fraction by a 3 and a 2 (or by a 6, if you’re feeling efficient). We need all of those factors to be represented in the LCD, but we only need the duplicates to be represented once. What I like to do is rewrite the last lines of the factor trees one on top of the other, to make it easier to see the overlap:

12 =        3 x 2 x 2
18 = 3 x 3 x 2

So that 3 x 2 of overlap we only need once. So let’s rewrite out the prime factors that we need in our LCD:

LCD = 3 x 3 x 2 x 2

There. We have two 3’s, so the 18 will be happy, and two 2’s, so the 12 will be happy. What does that come out to?

LCD = 3 x 3 x 2 x 2 = 36

So our target number is 36. Now, let’s go back to our original problem:

 3 + 4
12  18

Now, remember that any number divided by itself is one? Good. And remember that one of the definitions of a fraction is a division problem you don’t want to do yet? Great. So we can multiply either of these fractions by a new fraction composed of a number over itself, and not change the actual value of the number. We’re simply rewriting it in a form that’s more useful to us. It’s basically the backwards application of the process for reducing a fraction. Only this time we’re not reducing; we’re making it more complex.

We’re going to do this in two separate cases. Just look at the first fraction for a moment. What number do we need to multiply the 12 by to get our target of 36?

If you’re stumped, just go back to the prime factor makeup of our target number and figure out what pieces are missing from the 12:

12 =           3 x 2 x 2
LCD = 3 x 3 x 2 x 2

So we’re missing a 3. So multiply that first fraction by 3/3, which, remember, is just a fancy way of writing “1”, so we can do that without messing up the value of the fraction:

3    x    3 =  9
12        3     36

Okay, good sign. Our denominator is 36, which was our target. Let’s take the second case. Look at the second fraction:

4
18

18 =     3 x 3 x 2
LCD = 3 x 3 x 2 x 2

So we’re missing a 2 here. Multiply by 2/2, which is just a fancy way of writing “1”:

4   x   2 = 8
18      2    36

Excellent! Hang on, we’re almost done. Now, what we’ve essentially found here are equivalent fractions for each of our originals that just happen to have the same denominator. Well, they don’t really “just happen” to have them; we did that on purpose to make our lives easier. So now we just plug these new forms of each fraction into the original problem:

9  +  8
36    36

And we have our pies subdivided into the same size pieces! Now we can add them normally to figure out if we have less than a whole pie’s worth of leftovers:

9   +   8 =    17
36    36       36

17 is less than 36, so we have less than a whole pie. We can safely put them both in the same pan without overflow.

Now if I could just figure out how to fit that turkey in the fridge…

Four years ago, I posted this math trick on my blog.  Take a look at it, and at the end I’ll show you why it works!

~

Let’s play a game. I’m going to let you make up a math problem, and I will be able to tell you the answer from here. I can’t see what you’re doing, I’m not even in the same room as you, but I will still be able to tell you the correct answer.

Trust me. I’m a professional. Ready?

Okay. First, pick a number. It can be any number you wish, large or small. Now add 5 to that number. Got it? Okay, now double your new number (multiply by 2). Alright, now subtract 4 from the double.

Next, divide your new number by 2. Now, finally, subtract your original number from this new quotient. Got it? Okay. Here comes the cool part. Ready?

The answer is 3. Nifty, huh? What’s that? How’d I do it? Oh, magic.

Okay, okay, it’s not magic. The answer will always be 3, no matter what number you pick. Let’s illustrate this by writing it out as an algebraic expression.

Pick a number, any number. Since your number could be anything and is therefore a variable, we’ll call it b.

Add 5.

b + 5

Double that.

2(b + 5)

Subtract 4.

2(b + 5) – 4

Divide by 2.

[2(b + 5) – 4] / 2

Now subtract your original number.

([2(b + 5) – 4] / 2) — b

Okay, so let’s simplify this expression and see what we get.

([2(b + 5) – 4] / 2) — b

Let’s get that fraction out of there. Divide each term in the numerator by 2.

(b + 5) – 2 – b

That’s better. Now simplify that.

b – b + 5 – 2

5 – 2

3

See? It doesn’t matter what number you pick, because the variable cancels itself out at the end. The answer is always 3. Now, go forth and amaze your friends!

~

This game is a perfect example of the concept of inverse operations. Inverse operations are operations that cancel each other out; what I sometimes refer to as “undoing” each other. Addition undoes subtraction and vice versa, multiplication undoes division. Early in the problem you double your mystery number, and then later on you divide it by two. Those two actions cancel each other out – one makes the number larger and the other shrinks it back down.

In an algebraic equation, you can effectively move a term from one side of the equals sign to the other by performing the inverse operation to both sides. Y = x + 5 becomes y – 5 = x, which can tell you the value of x instead of y. Algebra, at its heart, is the process of using these inverse operations to rewrite an equation so that it tells you the piece of information you want to know.

Suppose I place you at one end of a long, empty room. Your task is to get to the door at the other end of the room. Simple, right? But what if I told you that this simple task is actually mathematically impossible?

Think about it – in order to traverse the whole room, you first have to get to the halfway point, right? You’ll have to travel one-half of the way there. And before you can get to that halfway point, you have to travel one-quarter of the way there (halfway to the halfway point). And before you can get to the one-quarter point, you have to travel one-eighth of the way there (halfway to the quarter-way point). Since you have to go half of each distance before you can go the full distance, you’ll never actually get anywhere. The task requires an infinite number of steps, and you can never complete an infinite number of steps since there will always be another one. Furthermore, in order to even start your journey you would need to travel a specific distance, and even the smallest of specific distances can be divided in half, giving you another step before that one. So you’d have to traverse infinity in order to go anywhere.

Make your brain hurt? Don’t be ashamed; it’s supposed to.

In fact, this is a very famous thought experiment called Zeno’s paradox. Zeno’s conclusion was that all motion must be an illusion, since travel over any finite distance can never be either completed or begun.

In practice, of course, this is not really a paradox – at some point your remaining distance will be so small that you cannot practically traverse half of it. Perhaps it would be down to a distance that is smaller than the length of your foot, so you are in effect already standing at both ends. The point of Zeno’s paradox, though, is to illustrate the elusive nature of mathematical infinity.

Infinity tends to cause problems for math students when they expect it to behave like a number. The truth is, infinity is a slippery concept, one that can only really be comprehended obliquely because if you ever try to stare directly at it it will squirm away. The way to understand infinity is to acknowledge that it is a concept that theoretically exists, but that you will never personally see or pin down. Very much like the imaginary number i, infinity has a definition but it cannot be evaluated as a number. This is why we can only talk about a series approaching infinity – we cannot say that it reaches infinity, because infinity cannot ever really be arrived at. Wherever you are on the number line, infinity is always out of your reach.