If you are having difficulty with algebra, these pages may be of some help because they offer different sorts of explanations, in some sense more "psychologically complete", than are usually found in algebra texts.
It is my belief that the way algebra is typically presented to students leaves out some ideas and explanations that are helpful, even sometimes necessary, for them to be able to do algebra well and to have a "feel" for it.
There are at least three different kinds of things taught in algebra courses: (1) language conventions, (2) logical numerical manipulations using those conventions, and (3) deducing answers to problems by using the conventions and the logical manipulations of them. I will explain as I go. But it is important for students to keep in mind whether in a given lesson they are supposed to be learning a convention, a manipulation, or a way of solving problems by using conventions and manipulations.
It is also my belief that school "culture" is such that, even contrary to good teachers' warnings, students will often think they are simply supposed to memorize formulas and recipes in math, rather than (also) understand them. Such memorization becomes a problem in courses, such as algebra, where understanding is at least as important as specific knowledge.
This is a two-fold problem. (1) Teachers need to try to give useful and helpful explanations -- and they need to be aware of as many typical student misunderstandings and confusions as they can; and teachers need to constantly try to monitor students for confusion and misunderstandings about what has been presented; waiting until there are test results is often too late. But also, (2) students need to know that THEY are the ones who will have to ultimately make the material make sense to them, and that they need to keep trying until it does. They may have to consult others, find a different book, or just sit down and think about the material, if they cannot understand their teacher's explanation about some aspect of algebra or other. There is simply no guaranty that any particular explanation will provide automatic understanding. Understanding requires reflective thinking of one's own. Explanations often are only a help to such thinking; and what serves as a great explanation for one student may not be helpful at all to another.
(My own first difficulty in "pre"-algebra was not understanding what letters such as "x" had to do with anything, and why letters were chosen to represent quantities at all, or how you worked with them when you had them. I vividly remember when the light dawned on me about this particular lack of understanding, in part because I still do not know why or how the teacher's particular explanation "worked" on me. She was saying that doing algebra was like unwrapping a package in the reverse way it was wrapped to begin with. It may be that I figured out what I needed to know while she was talking instead of because of what she specifically said, or it may be that what she said had some sort of meaning to me subconsciously somehow. I don't know, since "unwrapping" is not the way I see (or even then saw) algebra. But what follows are explanations of the sort that seem to me the most meaningful about some aspects of algebra many students typically have trouble with. Further explanations of other aspects of algebra can be found at A supplemental introduction to the first chapter of an algebra book and at Rate, Time Problems.)
The following question was asked on the Math-Help forum. It is typical of the kinds of problems had by students who don't really understand in general "what is going on" in algebra -- why you do certain manipulations of formulas, or how you choose which manipulations to do. Particular algebraic manipulations do not make sense to them because they don't have a general sense of what algebra is about, or what the point of the manipulations is. After I give the response to this question, a response which will include both general and specific problem-solving ideas, I will make come comments about how a typical algebra course is structured, and why it is structured that way.
I have a big exam Monday in algebra and I have
no idea how to do linear equations! Can someone please help quick? Her
are examples of a few that I am having problems with.
3(3 - 4x) + 30=5x - 2(6x-7) and 5x²-[2(2x²+3)]-3=x²-9
I am also having a few problems with this:
x + 3 + 2x = 5 + x + 8 (5)
I am supposed to figure out if 5 is the answer,
and tell how I got the answer.
My response:
It looks to me from this last problem that you perhaps don't have an
UNDERSTANDING of what doing algebra with equations is all about, which
makes doing any problems a bit difficult. But let's see what we can do
for you here. The following may be too much for you to absorb before your
test, though I hope not; but I think it is stuff you will need to know
for future tests as well, so maybe it will help you for them even if it
is too much for tomorrow's test.
Take the last problem first. Do you understand that if "five is the answer," that simply means that IF YOU WERE TO SUBSTITUTE 5 AS THE VALUE EVERYWHERE THERE IS AN X, THE STATEMENT THAT THE LEFT SIDE OF THE EQUATION EQUALS THE RIGHT SIDE WOULD BE TRUE? If 5 is NOT the solution, then when you make the substitution, the statement will not be true that the left side of the equation is equal to the right side.
Take a simple case first: X = 27 - 4
There is only one number X can be for this statement to be true; 23,
right? So 19 would NOT be a solution to this equation; that is, X cannot
equal 19 and the statement still be true that the left side equals the
right side.
Now make it just a bit harder: X + 3 = 27 -
4
There is still only one number this can work for but it is a different
number, because now we know that it is not X that equals 23, but some number
that, when you add 3 to it, gives you 23. So what number gives you 23 when
you add 3 to it? 20, right? That means X must be 20 in that equation.
It gets a little harder when you start putting X's on both sides of the equations and add or subtract some multiplications and divisions, etc., but the idea of what is going on is the same.
So if we look at the equation you gave last: if X does equal 5 in the equation x + 3 + 2x = 5 + x + 8, then that would mean 5 + 3 + 10 should equal 5 + 5 + 8. Does it? If it does, then X does equal 5; if it doesn't, then X cannot equal five.
If we thought X might be some other number, such as 2, then we would have to replace X with 2 to check. Would 2 + 3 + 4 = 5 + 2 + 8? If not, then X cannot equal 2 and the above equation still be true. All the "X" does is to tell you which number it is you don't know, and the problem is something like: I have a bag with some number of cookies in it; and if we add three more cookies to the bag and then add twice as many cookies as we had in the bag to begin with we get the same number of cookies as if we had taken the bag and added 5 cookies and then 8 more. There is only one number that could be in the bag to begin with that would make this statement be true; and that number is what X represents.
Now, of course, this sounds like a silly way to tell someone you had
a bag of five cookies, but they make up silly
problems like this so you can have practice learning the things you need
to in order to learn how to solve real problems where you don't know a
missing number that you need to figure out. Like in my business,
I need to enter in my books how much people paid me for photographs and
how much they paid for the sales tax that I have to turn in to the government.
Sometimes all I have is a check to go by that has the total amount, so
I need to figure out how much of it was tax and how much of it was for
the photography itself. I work where I must charge 8% sales tax. If I get
a check for $37.80 I have to compute how much the photography was and how
much the tax was. If I let X represent the price of the photography, then
I know that:
X + .08X = 37.80, since that means
that the price of the photograph plus 8% of the price of the photograph
is equal to the amount on the check. And I thus have an equation I can
solve to let me know which part of the $37.80 is for tax and which part
is for me.
Back to your equations. The idea is that you sometimes need to "multiply out" factors, and sometimes need to factor products into components in order to be able to get the X's and the numbers to such a point that you can see what the X is. If I told you that 3X = 9, you would know right off that X = 3, right? Well, there are strategies for getting the X's where you can tell what they are supposed to be, when you can't see it right off or any other way.
Let's go back to your last equation: x + 3 + 2x = 5 + x + 8
Normally, what you need to do is to find out how many X's you have altogether and what the total quantity is that those X's altogether give you. And to do that, you normally want to try to get all the X's on one side of the equal sign and all the quantities that don't have an X in them on the other, so you can find out that, say, 3X = 9 Well, in your equation, you have 3X's and a quantity 3 on the left side, and on the right side you have 1X and 13. So we need to try to get the X's to one side and the numbers that don't have X's to the other.
You have to understand that (when you see no other way to solve the
problem) the most important concept, or principle, or tool is:
Whenever you have any equation, where you
are saying
the left side = the right side you can always change either side of the equation by any amount you want to AS LONG AS YOU CHANGE THE OTHER SIDE BY THE EXACT SAME AMOUNT. That way, both sides will still be equal to each other even though both will be different from what you started with. There is a purpose for doing this with equations. |
(That is, if someone gives you and me the same amount of money, we can
know that no matter how much money that is -- even if we don't know how
much
it is -- that if we each double our money, we will still have the same
amount as each other. And we know that if we each triple our money and
then lose half of it and then add $5 to that, we still will have the same
amount as each other -- whatever that will be. No matter what you do with
your money, you can know that IF I do the same thing with mine, we will
still end up with the same amount of money as each other, no matter how
much or how little that will be.)
And, unless we see some other way to figure out the unknown quantities,
such as "X", we use this principle over and over in almost every problem
in order to get the X's on one side and the nonX quantities on the other.
And hopefully, then, we can get to a point where we can tell what X must
be equal to, since if you can get any equation into the form of
so-many X's = a kabillion,
you can figure out what ONE X is by dividing a kabillion by however
many "so-many" is.
The difficult part is to figure out which changes to make to both sides that will be useful and helpful for you to figure out what the variables (or "unknown quantities" or "unknowns") are. Some changes will be more useful than others. Many won't be helpful at all. Normally, but not always, you want to try to do whatever manipulations will help you get down to a quantity that is equal to ONE variable, that is, one X or one Y or whatever letter(s) you chose to represent the quantity you are trying to figure out. But it is not always easy to see which manipulations might help you do even that. Understanding, combined with practice, help; but solving any particular (new) problem may also take some trial and error, a flash of insight, or some luck. (Return to previous part of text if you came here from the link.) |
So, back to your equation: x
+ 3 + 2x = 5 + x + 8
we added all the X's up and all the numbers up on each side of the
equation, and we found out that
3X + 3
= X + 13. Well, we can subtract the X on the right side in order
to "get rid" of it AS LONG AS WE ALSO subtract X from the left side. Hence,
we get: 2X + 3 = 13 (Or, in intuitive
terms, you might see that if we DON'T add the X to the 13 on the right
side, that is the same as not having added one of the X's in on the left
side to begin with, which would have just given us 2 X's on that side.)
If you still can't see how much X must be for this to be true, we can use the same principle, this time subtracting 3 from each side of the equation, so that we are left only with 2X = 10 (which just says that some number multiplied by 2 equals 10). If these were big numbers and you still couldn't see it, then we would divide both sides by what we are multiplying the X by (in this case "2") in order to see how much 1 X is; 5. If we had 33X = 9999, you would divide both sides by 33 in order to get what one X is.
Do you get the idea here?
Now we go back to the first problems: Take 3(3 - 4x) + 30=5x - 2(6x-7) first.
They have added an extra wrinkle to this one, because instead of telling you how many X's you have on each side, they have told you that you have some X's that get manipulated by subtractions and multiplications, etc.
So the first thing to do, usually, but not always (I'll give a counter-example at the end of all this) is "multiply out" the quantities, giving: 9 - 12X + 30 = 5X - 12X + 14
(Time out, in case you need it: do you see why it is PLUS 14 here, instead of minus 14? If not, click HERE.)
Back to working out the above equation. We had it to:
9 - 12X + 30 = 5X - 12X + 14
(There are at least two ways we could proceed from here. I am going to go through the "standard" way in the text and put the non-standard way HERE.)
Now it is just like the one we did before where we had some X's and some non-X quantities mixed together on each side of the equal sign. In this case we have:
39 - 12X = -7X +14
At this point we start trying to do things equally to BOTH sides of
the equation in order to try to end up with all the X's on one side and
all the non-X's on the other side. So we can either try to get the non-X's
on the right and the X's on the left, or we can try to get the non-X's
on the left and the numbers on the right. It doesn't really matter from
a technical standpoint which way you choose, but from a psychological standpoint,
it is usually easier to work with positive numbers than negative numbers,
so instead of subtracting 39 from both sides and ending up with -25 on
the right side, it is psychologically easier to subtract 14 from both sides
and end up with 25 on the left: 25 - 12X = -7X
Then, if you add 12X to both sides, you end up with 25 = 5X, and since
that just means five times some quantity is equal to 25, you know the quantity
must be 5. So X = 5.
(If we had gone "the other way" and started out
by subtracting 39 from both sides of the above equation, we would have
got -12X = - 7X - 25, and then to get the X's all on the left, we would
have had to add 7X to both sides, giving us -5X = -25, which will still
mean that X = 5, but it is usually psychologically more difficult to work
with and to see that way.)
Now, it is important to make sure you did it right and got the right answer. To make sure you did it right, you go back to your original equation and put 5 back in wherever there is an X, and see if it comes out true:
The equation was: 3(3 - 4x) + 30=5x - 2(6x-7)
which gives us 3(3-20) + 30 = 25 - 2(30-7)
So, does 3(-17) + 30 equal 25 - 2(23)?
Does -51 + 30 equal 25 - 46?
Going on to your second problem:
5x²-[2(2x²+3)]-3=x²-9
This one is something of a trick AT THE LEVEL YOUR ARE STUDYING because when you have Xsquared = something, you get TWO answers: a plus and a minus answer; e.g., if X^{2} = 9, X can be either plus 3 or negative 3, since (-3)(-3) = 9 just as (3)(3) does. It will turn out that when you graph equations that have squares (in their most simplified version), you don't get straight lines, so they are not straight-line or "linear" equations. So the trick here must be that either the X^{2} will all disappear, leaving you with a linear equation or something else will happen where you end up with something like X^{2}= 36 and you know X will be either 6 or -6, or something else weird will happen (and it is this last thing that actually happens here -- your book or your teacher must have a sense of humor, perhaps a slightly sadistic one?). So, let's see what happens when we do all the operations in order to get to a simplified statement of the equation: 5x²-[2(2x²+3)]-3=x²-9
Multiply it all out and you get:
5X^{2} - [4X^{2} + 6] - 3 = X^{2}- 9
X^{2 }- 6 - 3 = X^{2 }- 9
which is just to say that X^{2 }- 9 = X^{2 - }9
or
9=9
In other words, this equation will be true for every value of X since the equation is something like: X + 12 = X + 12, which is always true, no matter what X is. Weird problem to give you.
Finally, above I said that you USUALLY need to multiply out all the factors they give you, BUT you don't have to ALWAYS multiply factors out, if you happen to see that a factor is duplicated somewhere, or duplicated on each side of the equation in a way you can get rid of it.
Suppose we had: 3(2x - 5) + x = 4 + 3(2x - 5)
Since there is a 3(2x - 5) on both sides of the equation, we can subtract that quantity from both sides without having to figure out WHAT it is. That would give us x = 4 without having to do all the work of multiplying and regrouping and everything.
Or, if you had 6(2x - 5) divided by 3(2x - 5), you can know that is equal to 2, no matter what (2x - 5) equals.
The Structure of Algebra Courses Typically
Typically students are taught a number of principles and manipulations that seem to many of them to have nothing to do with anything. Then they are given practice in using those principles or performing those manipulations -- principles and manipulations which may or may not make any sense to them. For example, they are taught about association and commutation, and given practice "simplifying" or "multiplying out" expressions such as a(b + c) or (a +b)(3c - 5d). They are taught the rules for "order of operations" and then given practice calculating expressions that are written without parentheses. Or they are taught to factor expressions like ax + bx into x(a+b). Or they are taught you can do the same things to both sides of an equation, as long as you do the same thing to each side. Then they are given equations to solve, which seem to use a bunch of those manipulations or principles. And finally they are given problems in words which seem to have something to do with equations and manipulations. |
I write the above the way I did because much of algebra seems rather arbitrary to students who do not understand the individual manipulations and principles (in algebra overall, or in a particular chapter or unit) or who do not understand their point. Some of these students will not be able to remember the manipulations well enough to do them very well; others will be able to use them rather mechanically to solve problems of a type they have been trained to solve, even though they don't really understand why one goes about those particular mechanics other than that they give you the answer the teacher likes. For both of these types of students, new problems will be particularly difficult.
What is actually occurring is that it turns out there are certain (kinds of) logical, sensible, reasonable principles and manipulations that tend to be useful in solving certain kinds of problems that are the typical algebra problems, or, to put it perhaps better -- there are certain (kinds of) logical principles and manipulations that tend to be useful in solving the sorts of problems that algebra lends itself to solving. That is why students are taught these particular manipulations and ideas. It is important for students to see the logic and the sense in the principles they are taught, not just to learn the rules as some sort of arbitrary recipes. The principles themselves are logical, not arbitrary, although the way they are stated, the order in which they are introduced, and the particular ones chosen for a particular book or chapter may have been quite different.
One of the important aspects taught in algebra is precision of expression, so that one can learn to express in numerical, usable form ideas or problems that normally occur first in words. The example I gave above is one such case, where I was entering checks into a ledger and needed to figure out how much of the amount on the check was for sales tax and how much was the price of the object. Another case might occur in something like baseball where a batter might want to know how many hits he may need in his next 10 at bats to raise his average to a certain level. Or you might want to know how much money you need to take on a certain car trip in order to pay cash for gasoline. Or you might want to compute how much interest you paid on your mortgage last year versus how much of what you paid went for principle. Solving these problems generally requires your being able to express the problem in numbers related to each other in some way, and then knowing how to go about manipulating the numbers.
Some of the concepts apply not just to math, but to language in words. I was driving home tonight listening to a news program on the radio in which they were talking about celebrating Dr. Seuss's birthday in many schools, and in one school, a chef had prepared "green eggs and ham" in honor of the book by that name. Well, I was familiar with the book, and had read it many times to my own children. However, I was surprised when the news reporter asked the chef, after he had explained how he made the eggs green, how he had made the ham green. I was surprised because it never occurred to me that the ham was supposed to be green. I thought Green Eggs and Ham meant Ham and Green Eggs; I thought the "Green" went just with the "Eggs", not with the ham too, or not with the Eggs-And-The-Ham. To put it in math-like terms, I thought the title and story were about (Green Eggs) and Ham, not about Green (Eggs and Ham) or Green Eggs and Green Ham. When I mentioned it to my younger daughter, she put another interpretation to it -- Green Ham-and-Eggs, that is a greenish mixture of ham and eggs, not just Green Ham with Green Eggs. So now I am not sure at all what Dr. Seuss really meant, because he didn't use parentheses in the story or the book title and he didn't write about the individual components....
So part of algebra is learning the language of how to write down consistently precisely what you mean when you are trying to express or represent something numerically; it is about learning a precise language. As you learn to do this, you should be thinking about what a given numerical expression might mean if you write it one way versus another. If you do that, then numerical expressions will begin to take on meaning for you and not just be a bunch of symbols you are trying to manipulate. And if you understand their meaning, you will be able to figure out ways to manipulate them when you cannot remember by rote how to do something. For example, if you go to add 1/a to 1/b, you might be tempted to think it will come out to 2/(a+b), but if you understand that 1/a is a fraction, and that 1/b is a fraction with a different denominator, you can check to see whether you can add fractions that way by looking at fractions you know, say adding 1/2 to 1/4. You know that will be 3/4, so the question is whether your inclined way to do it will also give you three fourths. However, you will see that it gives you 2/6, which is 1/3 and is not anything near 3/4. That way you know your inclined way of adding 1/a and 1/b is not right. If you weren't thinking about what 1/a and 1/b meant, you might have gone ahead and just made the erroneous combination. At a more complex level, suppose you are doing one of those problems where it takes one guy 3 days to do a job by himself, another guy 2 days, and a third guy a day and a half, and they want to know how long it will take them all working together to do the job. If you get an answer of more than a day and a half, you know something is wrong, because that means that when the fastest guy has help, it takes him longer to do the job than it does when he is working alone. (Now, of course, some people DO slow you down when they try to help you, but that is not the intent of problems like these.) So you not only want to think about your problems and your answers in numerical terms, but what those numerical terms mean in your own language.
Another tool that tends to crop up repeatedly in working problems in
an algebraic way is multiplication across or through parentheses -- expressions
of the sort (a+b)(c+d), or even more complicated by having more components
or more multipliers. You need to be able to do these sorts of multiplications
quite often in order to be able to get to the specific quantities of a
certain variable (or unknown) and the quantities of non-variables. E.g.,
if you have some problem that starts out being expressed as
8(x + 3) - 4(x - 21)=5(x-1) + 3(4x - 2),
you will likely need to multiply all that out in order to figure out
how many X's equal how much.
But, on the other hand, there will be times you will want to be able
to factor expressions into components in order to work with them. For example,
if you have (X^{2} - 49)/(X + 7) = 14,
you can figure out pretty easily that X = 21 because any expression of
the form (A^{2} - B^{2}) can be factored, as you normally
would be taught in an algebra class, into (A + B)(A - B), which means that
the above expression will factor into (X + 7)(X
- 7)/(X + 7) = 14
And that means that you can then divide the (X + 7) in the numerator
by the (X + 7) in the denominator and end up with just X - 7 = 14, which
means X must be 21. And if you work out (21^{2} - 49)/(21 + 7),
you will get (441 - 49)/(28), which is 392/28, which is 14, as the original
equation stated. So by factoring you made it easy to calculate something
that would otherwise be difficult to see or figure out.
Algebra courses also often teach about the relationship between graphs, or certain kinds of lines and curves and shapes (that appear on graphs), and numerical expressions or representations of those things.
I am not sure whether it is true or not, but it may be helpful to think of many branches of math as being potentially able to represent or express certain important characteristics of phenomena of all kinds in numerical and/or logical terms.
Basically what you want to do when studying algebra is to make sure you understand 1) what expressions, equations, formulas, and manipulations really represent and why they are able to represent what they do, and to make sure you understand 2) how to do the sorts of manipulations they give you, AND how and why those manipulations work. And you want to ask your teacher how those representations and manipulations will likely be useful later when you get to actually working real problems, or at least the sorts of problems expressed in ordinary language in the algebra class. Then you also want to have practiced sufficient representations and manipulations sufficiently to be able, not only to understand them but, to do them fairly quickly and automatically.
Otherwise what will tend to happen to you in an algebra class is that you will just try to memorize sequences of equations and their manipulations that don't really make any sense to you, but which you can memorize -- UNTIL you get too many to be able to retain or UNTIL you have to figure out WHICH MANIPULATION is the one you need at a particular time. It is one thing, for example, to be able to work out problems involving (A^{2} - B^{2}) when they come at the end of a chapter teaching about (A^{2} - B^{2}); it is quite another thing to be able to recognize something in that form and realize the form can be useful to you three chapters later in the book when you haven't been working specifically with that form and aren't looking just for it.
Part of the reason students tend to memorize aspects of algebra that they instead ought to be trying to understand logically is that they wasted time at the beginning of algebra trying to understand "conventions" that did just need to be memorized instead of being understood logically. How mathematicians understand the use of parentheses (or their absence) is a matter of convention that just has to be learned by memory, but the consequences of that choice have to be understood logically as you practice doing different manipulations, such as adding fractions with different denominators. For example, you cannot logically add 2/3(a + b) to 3/3a + b to get 5/3a + b, because the denominators are not the same even though they look similar. Suppose, for example, that a = 4 and b = 6. The first fraction then is 2/30 and the second fraction is 3/18. And their sum is neither 5/18 nor 5/30.
So, if you have a problem such as
2) Learn to do various kinds of calculations using the conventions. These are a matter of logic and understanding, prior to practicing them to help you use them better. 3) Learn to solve equations, normally by using the logical manipulations and calculations in a logical and creative manner to isolate the unknown variable on one side of the equation and its eqivalent on the other. This is a matter of insight, logic, creativity, and sometimes luck. But insight, creativity, and luck can be improved many times by practice and understanding. |
There is an analogy to much of this in learning computer languages, or even in building web pages with HTML codes. It is one thing to learn HTML codes or functions you can make the computer do in a given programming language, but it is quite another to see how you can build elaborate web pages or create very complex functions and utilities by simply utilizing a few simple codes or functions in ingenious combinations.
Rick Garlikov (Rick@Garlikov.com)
There are two ways to get this -- the short way
of just following a rule (either the rule that when you multiply a negative
by a negative, you get a positive; or the rule that when you subtract a
negative number, that is the same thing as adding the number), and the
longer way where you understand what you are doing -- which is to realize
that in the above case you are subtracting (double) a-number-that-is- going-to-be-made-smaller-by
7-first. Let's look at an example in numbers first:
Suppose you have 20 things and someone thinks he wants to buy 9 of them. If he does, you would have 11 left, since 20 - 9 would be 11. Now suppose someone else comes along and wants to buy some of those things but for some reason he says "I want 7 less than the last guy bought". Well, he wants 9 - 7 then, or 2. There are at least two ways you could figure out how many you will have left after this purchase:
1) you could just say, he is buying 2, and since I have 11, I will have 11 - 2 left, or 9. Or
2) you COULD say (but you probably wouldn't) that
what you will have left will be 11 - (9 - 7), since
(9 - 7) is in numbers what the guy told you in
words that he wanted -- seven less than the last guy bought. Hence, you
are subtracting 9 from the 11, BUT you are adding back the seven the guy
did not want to take from you. In essence he is subtracting seven less
than the first guy, so if you subtract the same amount the first guy took,
you will be subtracting 7 too many.
Either way it should come out the same, because either way you figure it out, it will still be that you will have 9 things left.
And since, in the equation they gave you they double the quantity after they subtract something from it, it would be like the second guy's saying he wanted to have twice as many things as seven less than the other guy, which, of course is a stupid way to talk. But suppose he and his brother each have the same number of kids and that he and his brother are trying to keep up with the Joneses who have seven more kids than each of them has. So he wants to buy the same things for his kids and for his brother's kids that the Joneses buy. So he goes everywhere Jones does, and any time Jones buys something for each of his kids, he wants to buy the same number of things for his own kids and for his brother's kids. So he always tells the sales person, "whatever that guy bought, I want to buy seven fewer -- but after you figure out how much that is, I need you to double it, since I am buying the same amount for my brother as I am buying for myself." This is represented by 2(Jones' purchase - 7), and it will come out to be twice what Jones purchased, minus 14. So, if you were keeping a running inventory of what you had left, and you started with, say, 100 of these things, Jones would leave you with an inventory of (100 - Jones'purchase) and the next guy's purchase would leave you with that amount minus the following amount:
2(Jones'purchase - 7)
and if you play around with some numbers, you
will see that always ends up where the second purchase is 14 less than
the first purchase, and therefore leaves you with 14 more items left in
your inventory than if the second guy had simply bought twice what Jones
did. (Return to the main text.)
The equation we have arrived at is 9 - 12X
+ 30 = 5X - 12X + 14
Notice that 12X is subtracted on both sides of the equal sign. This
means that the same amount is being taken away from both sides and therefore
the rest of what is left on either side will still be equal. So instead
of adding 12X to both sides, we could actually just eliminate subtracting
12X from both sides, and either way, arrive at: 9 + 30 = 5X + 14, which
is to say that 39 = 5X + 14. So if we didn't add the 14 to the 5X, we would
have 14 less than the 39, which would be 25; meaning that 25 = 5X, which
is just to say that some quantity multiplied by 5 will give 25. And, of
course, that quantity is 5. So X = 5.
(return to the text)
Factoring & "Multiplying Out"
By "multiplying out" an expression, I simply mean taking an expression such as a(x + y) and doing the multiplication which is indicated so that you get ax + ay. Or, if we were to take an expression such as 5(15x - 4y), you would get 75x - 20y. Or if we had (3x + 2)(4x - 5), that would give us, when we "multiply it out" 12x^{2} - 15x + 8x - 10 or (combining the "x's") 12x^{2} - 7x - 10.
"Factoring" just means going in the opposite direction; that is, dividing
a quantity into factors or components which, when multiplied together,
would give you that quantity. The factors of 10 then would be 5 and 2.
Factors of ax + ay would be a
and (x + y) so that when you "factor"
ax + ay, you get
a(x + y). From the paragraph above,
if you factor 12x^{2} - 7x - 10,
you get (3x + 2)(4x - 5).
Multiplying out expressions is pretty much just a mechanical procedure
where you multiply each of the terms times the other terms and then combine
the "like" or similar terms -- numerical quantities, quantities with the
same variables to the same power, etc.; but factoring takes some insight,
and often a great deal of luck in seeing combinations. For example, it
is not easy to see that the above expression 12x^{2}
- 7x - 10 even factors, let alone see what the factors are.
If you have been doing "factoring problems" in a text book chapter, you
start to see patterns that the author has begun to use, or that are stereotypical
in textbooks, but those are not usually in your mind when you are working
real problems later and get into an expression that is factorable but not
obviously so. (return to text)
Sometimes you can do a problem without having to go through lots of
manipulations because you can see the answer right off. Physicist Richard
Feynman, when he was in high school, answered a difficult algebra problem
immediately during a math tournament, without doing ANY algebra or math
at all. It was something of the sort where a rowing team during practice
is rowing upstream against a current that is moving 4 mph relative to the
shore. They are making progress relative to the shore at the rate
of 1.5 mph. The hat of the guy in the back of the boat falls off into the
river without his realizing it, and it floats downstream with the current.
After 15 minutes they realize they have lost the hat, and immediately begin
rowing back downstream to retrieve the hat. If they row with the same strength
or power they were rowing upstream, now that they have the current going
with them, how long will it take them to retrieve the hat? Feynman saw
immediately that it was 15 minutes because he realized that all the rates
relative to the shore were irrelevant to the result, just as when you drive
west on an open stretch of level road for an hour and then drive back at
the same rate, it doesn't take you longer to go one direction rather than
the other, even though the earth is turning eastward at the rate of approximately
1000 mph (near the equator).
I found out, through a different, counter-intuitive, "trick" problem one time that not only is the circumference of any circle six and a quarter times its radius, but that when you increase the radius of any circle any amount, you thereby increase the size of its circumference by roughly 6.25 times that amount. Therefore, if you tell me that you added 10 inches to the radius of a dime, or 10 inches to the radius of the universe (assuming it is round, which may not be true, of course), I know that you have increased the circumference of each by the same amount -- roughly 62.5 inches.
Of course, in an algebra course, teachers won't tend to accept
such reasoning, and want to see a "mathematical proof" -- which can be
given for both of these cases; but I am simply saying that in real life,
one does not necessarily need to solve what look like algebra problems
in an "algebraic way". The idea in real life is to solve problems however
you can. Algebraic manipulations -- particularly standard cookbook manipulations
-- are just one way to solve certain kinds of problems. Feynman himself
seemed to hold that the algorithmic or cookbook types of rules one learns
in algebra class are really just rules for people to follow who don't really
understand mathematical thinking. I wouldn't go quite that far, but I would
say that following such rules is only one way to solve many (apparently)
mathematical problems. (Return to text.)
These are what are usually called "word problems" because they are logical/numerical problems expressed in ordinary language. They are attempts at giving you some practice in solving problems the way they supposedly appear in real life, since real life problems don't appear already set up in formulas or equations.
However, unfortunately, real life problems
also don't appear formulated as structured as word problems do; and they
don't appear at the end of a unit so that you get an arbitrary, circumstantial,
clue to what is expected. (return to
text)
Languages are "conventions" because what words or phrases or marks on paper mean is what people decide they mean or what they happen to grow to mean. There are private conventions, such as secret codes, and public conventions such as ordinary languages. There are also private conventions for individuals. For example, if you are measuring and sawing wood, you might mark the length of the wood to be cut by putting your mark even with the end of your measuring tape, or you might put the mark just outside the measuring tape. Either way is fine, as long as you know when you go to cut the wood, which way you chose to make the mark. If you made your mark even with the end of the measuring tape but cut the wood inside of that mark because you thought you had made the mark on the outer edge of the tape, your wood will be a fraction shorter than you intended. So it is important to remember what your mark really means. You can do this by remembering how you do things each time you do them, even if you do them differently each time. Or, usually easier, you choose a way to do it every time and then do it that way each time, and there is less to remember. You have developed a convention for yourself.
In math and science, ambiguities of ordinary verbal language (such as in the "Green Ham and Eggs" case) are tried to be eliminated by developing and improving as necessary, precise symbols that have very specific meanings. This is intended to prevent confusion in communicating to others what you are doing. But it also helps you keep straight yourself what you are doing, by not having to remember each time what you meant by using a specific symbol.
Conventions have something of an arbitrary nature, however, though once developed they might have some sort of logic or psychological sense of appropriateness or seeming reasonableness of their own. They are arbitrary in that we could have used any symbol to designate what we want to designate. And in some cases how we use them is arbitrary. For example, in algebra, "3x + 1" is understood by convention to mean the quantity that is one more than the product of 3 times the variable "x", rather than the product of three times the number which is one larger than x. If "x" were 8, "3x + 1" is 25 in the way the convention or language of algebra is used because it is 3 times 8, which is 24 and then 1 is added to give 25. But if there were no such convention, someone might take it to be equivalent to 27, because they thought it meant 3 times the sum of eight and one, which would be 3 times 9.
In algebraic notation, that is to say that "3x + 1" just is designated to mean the same thing as "(3x) + 1" rather than "3(x + 1)". Without that convention "3x + 1" would be ambiguous and possibly mean different things to different people, or you might use it one way when setting up a problem and then mistakenly think it meant something else when working the problem -- as when you accidentally forget to keep the parentheses when going from one step to the next while solving a problem and then get the wrong answer.
In algebra, sometimes they state conventions in ways that make it sound like they are a conclusion to some sort of logic rather than merely an arbitrary convention or choice that was once made. This confuses students who are trying to understand "why" something is the way it is when, in fact, there is no logical reason, because it is simply a convention. For example, using the "x" below to mean simply the multiplication symbol (not an unknown variable) some teachers at the beginning of algebra will ask students what the answer is to something like
But conventions, once established, also have a resulting logic that
is not arbitrary. For example, because the English drive on the left
side of the road, and Americans drive on the right side of the road, the
way they each have to make right or left turns from two-way streets onto
two-way streets in their own countries is different. When Americans
turn right, they stay near the curb. When they turn left, they have to
"swing wide" to get to the right side of the street onto which they are
turning. The English have to do just the opposite, hugging the curb
lane to make left turns and swinging wide to the outside lane to make right
turns. That is a logical result of choosing which side of the road
to drive on, even though the choice itself had no logic but was simply
a matter of convention. Moreover, there is another logical result
of the convention. Although we are taught to look both ways before
crossing a street, we tend to look one way, toward the traffic approaching
on the curb side, as we step off the curb; and we look the other way as
we get to the center of the street. In America, one generally then
looks left before stepping off the curb, and right once one gets to the
middle of the street. In England, that can put you in front of a
bus, because the traffic next to the curb is coming from your right, not
your left; and the traffic on the far side of the center line is coming
from your left not your right. The English have the same problem in reverse
when they are in America. That is a very difficult habit to overcome when
visiting the other country and walking across streets. But the reason
you have to look the way you do as you cross a street is a logical result
of the side of the road the country chose to have drivers use. So
conventions which are arbitrary, nevertheless can have consequences that
logically depend on them. (Return to text.)