Here are some notes on several topics I have found students tend to have trouble with. I may add to this from time to time as I come across other common questions.
Factoring and Fractions (GCF/LCD)
May people have trouble with fractions. Be sure you spend enough time here so you really feel comfortable working with fractions. Here are some notes on finding the GCF and LCM/LCD, if you are having trouble with that:
There are several ways to do this, but I will just cover the method that works for very large numbers. (It also works for small numbers, though one of the other methods might be simpler for those.) First, you must find the prime factorization of both numbers. Once you have done that, you look at those factors and choose which ones (and what powers) will go together to find your GCF or LCM. Here is how you do that part:
Greatest Common Factor (GCF) - Choose only the factors that both numbers have. For those factors, choose the smallest power.
If you have:
24 = 2 x 2 x 2 x 3 = 2^3 x 3
80 = 2 x 2 x 2 x 2 x 5 = 2^4 x 5
So, to find the GREATEST common factor (GCF):
Both have a 2 in them; 24 has 3 of them and 80 has 4 of them (look at the powers of 2). We only choose factors that BOTH have.
So, the GCF is something common to BOTH of them, so that means the LEAST number of 2's, which is 3 of them. Since only one number has a 3 and only one number has a 5, they don't show up. The GCF is:
2^3 = 8 (which is a factor of 24 AND of 80)
Notice that 2^4 would NOT work (2^4 = 16, which is NOT a factor of 24).
Least Common Multiple (LCM): Take all the factors (from both numbers) and pick the largest powers for each.
Using the same numbers:
24 = 2 x 2 x 2 x 3 = 2^3 x 3
80 = 2 x 2 x 2 x 2 x 5 = 2^4 x 5
Now, to find the LEAST common multiple (LCM), we need ALL the factors (not just the common ones):
We need a 2, 3, and 5. However, to get EVERYTHING, we have to take the GREATEST number of times (biggest power) for each of these:
2^4 (biggest power of 2),
3 (only power of 3), and
5 (only power of 5)
So, the LCM is:
2^4 x 3 x 5 = 16 x 3 x 5 = 240
Signs in multiplying and dividing real numbers
One way of thinking about negative signs is to remember that two negative signs multiplied or divided always become a positive (we sometimes say they "cancel out"). If you are multiplying/dividing a bunch of numbers, it can also be handy to count the negative signs. If you have an odd number of negatives, the answer will be negative, but if you have an even number of negatives, the answer will be positive. So, for example:
(-2)(-3)(-1) = -6 since there are 3 negatives
(-1)(-2)(-3)(-4)= +24 since there are 4 negatives
Or, a more complicated example:
(-2)(-15) --------- = - 30/3 = -10 since there are 3 negative signs total 3(-1) (2 up top, 1 down bottom) Or: (-1)(-20) --------- = 20/10 = 2 since there are 4 negative signs total (-2)(-5) (2 up top, 2 down bottom)
One warning: This only works if you are multiplying and dividing only. If you throw in some addition or subtraction, these rules only work on the terms that are multiplied together (or divided). Basically, if you just follow your order of operations and take care of your multiplcation and division first, you shouldn't have any problems.ve any problems.
Another common problem: If you see the problem -5^2, many people tell me the answer is 25 (positive). However, the "order of operations" says that the exponent must be taken care of first, so:
-5^2 = - (5^2) = - 25
Compare that to (-5)^2, which does equal 25 (positive), since the parenthesis says the negative happens first:
(-5)^2 = -5 * -5 = 25
Algebraic rules and mathematical properties
Here are some notes I wrote in answer to a question about the different mathematical "properties" and how you know when to use them:
I hope you find at least some of this helpful.
Percentage appications and word problems
Here is an example of a common mistake (and the correct way) that I often see with percentages:
Thelma ordered some books over the internet. Her total bill was for $54.49, including shipping and handling. If she knows that the site charges a 6% fee to cover shipping and handling, how much did her books cost BEFORE the shipping and handling fee?
Since you want to know the cost of her books before shipping/handling (S/H), let:
x = cost of books without S/H
Now, you know that:
Total cost of order = (cost of books) + (shipping and handling)
Well, shipping and handling is 6% of the cost of her books:
shipping and handling = 6% of cost of books = 0.06 * x
(A reminder: * means "times", so 6% times the cost of books. Don't forget that you have to change 6% to a decimal 0.06 to use it in an equation.) Thus:
Total cost of order = (cost of books) + (shipping and handling)
= x + 0.06 x
But, we know the total cost is $54.49, so:
54.49 = x + 0.06 x
To solve this:
54.49 = x + 0.06 x = 1 x + 0.06 x
54.49 = 1.06 x
x = 54.49 / 1.06 = 51.41
(rounding off to 2 decimals)
Thus, her books alone cost $51.41.
A warning: Some people try to work this problem by the following method:
Since she paid 6% for shipping and handling, that means shipping and handling cost 6% of 54.49, so:
shipping and handling = 0.06 * 54.49 = 3.27 (rounded off)
Now, just subtract that from her total:
54.49 - 3.27 = 51.22
Now, at first, this sounds good. Unfortunately, it is wrong. Why?
The reason is that she is supposed to pay 6% of the ORIGINAL price for shipping and handling (if her books cost $10, she should pay 6% of 10, or 0.06 * 10 = $0.60, for shipping and handling). However, in the work above, we multiplied the 6% times the FINAL cost (since we didn't know the original cost), which is not the same number.
You can check to see that this doesn't work: If her books really cost $51.22, then she should pay:
6% of 51.22 = 0.06 * 51.22 = $3.07 for shipping and handling
So, her total cost would be: 51.22 + 3.07 = 54.29, which is wrong. Think through this carefully until you really understand it. This is an important point.
Another tip:
If the problem says "the sum of 2 numbers is 21", then that means you can always rewrite this as:
"one number = 21 minus the other number"
So, as an example, if you were to go to the store and buy two different brands of cereal and it cost you $6 total (in other words "the sum of the two prices was $6"), then you know that:
the price of one of them = 6 - the price of the other one.
Sometimes putting things in terms of real objects rather than just numbers helps.
Also, I sometimes people make the following mistake:
If the problem says "the sum of two numbers is 50", they write:
x + x = 50
This is not the "sum of two numbers", it is "one number plus itself" (because they used x for both). If you are adding two different things together, you can't use x for both of them; you can use x for one, but you have to find a formula for the other one.
Here is some general advice on solving "word problems":
Many people see a word problem, say "OH NO!!!! I HATE WORD PROBLEMS!!!!!!! I CAN'T DO WORD PROBLEMS. THEY'RE STUPID ANYWAY...", and give up. (Not surprisingly, they tend to not do well on them...) Now, I'm not one of those "Don't worry, just think happy thoughts." type of people, but I think you would find it helpful if you can defuse the panic and approach them a little more calmly. I understand the frustration; "word problems" can be hard. In addition to learning the basic math (which is often the easiest part), you also have to learn how to translate English sentences and words into math notation and then do some "problem solving" to come up with an equation you can solve. However, if you hang in there and really practice this a lot (whatever you do, don't avoid them and hope they will just go away - they won't), you will find they do get easier over time.
Try to break things down into smaller parts. Here are some other general suggestions:
I realize that these are all kind of general. If you are hoping for a few rules that will always tell you what to do with word problems, I'm afraid there aren't any. Learning to be good at working these is really a matter of practice and trying not to go whacko. It will get easier with time, but it does take time.
P.S.: Even though people often say they wish they didn't have to do word problems, you should remember that they are some of the most important and useful things you will learn in a math course. (Don't even ask me how often students say "What are we ever going to use this for in real life?") Learning to translate a real-life problem into a math problem you can solve can be very useful. I will be the first to admit that many of the problems you do aren't very realistic, however. This is mostly because "real-life problems" are usually more complicated, so we are starting you off with some simpler problems (even though it probably doesn't feel like it at times). Try to keep this in mind as you practice these.
Here is some more general advice on solving "word problems":
Another thing that can sometimes be helpful is to start grouping word problems together. For example, if you look back over the percentage problems, a common one went something like this: "The original price is increased/decreased by so much and the final price is $100. Find the original price." If you look back over all the percentage problems in the handout, participation problems, and test review, you will see that there are several like this. If you look at how you worked them, you will see that you worked them all pretty much the same way (several of these have been posted to the messageboards and I will be posting the solution to the participation problem). That means, if you run into a similar problem on the test, it will be worked in the same way. (Of course, there are other types of percentage problems, so be careful they really are the same.) Another example is: "The sum of two numbers is blah. One number plus blah, blah, blah..." All of these have some similarities to each other.
My point here isn't to try to "memorize" all the kinds of word problems; that won't work (there are way too many). However, if you can start recognizing "Hey, I've seem a problem just like this before", you can then think back to how you solved it the last time. That way you aren't starting from scratch all over again on each problem, but using what you have learned on other problems to help you. Now, this isn't easy, but it should be something you practice when doing your homework and studying.
As you practice this, it will become easier. If you go to the tutoring labs, you might ask them to help you figure out if a particular problem is "like" some of the ones you have done before.
Some reminders on solving inequalities:
Solving linear equalities is almost exactly like solving linear equations, except whenever you multiply or divide by a negative number, you have to remember to reverse the direction of the inequality. So, for example, if you were to solve the equation:
-3x + 5 = 11
-3x = 11 - 5
-3x = 6
x = 6 / -3
x = -2
However, if it were an inequality, you would make the following change:
-3x + 5 > 11
-3x > 11 - 5
-3x > 6
x < 6 / -3 <===== This step is different, since you are dividing by a negative (notice that I reversed the > into a < because of this).
x < -2
Notice, except for that one step, these problems are worked exactly the same. So, two bits of advice:
Slopes and Graphing Equations
If you are working with the slope to go from one point to another (without having to graph), if you know that the slope of a line is 7, well that is the same as 7/1. But remember that slope means:
slope = rise / run
or
slope = (change in the y direction) / (change in the x direction)
so
7/1 = (change in y) / (change in x)
so the change in the y direction is 7 (the "rise") and the change in the x direction is 1 (the "run").
So, if you start at the point (3,5), for example, you could find another point by changing your y value by 7 and your x value by 1:
new point = (3+1, 5+7) = (4,12)
If your slope was -2/5 and you started at the point (2,-3), then:
-2/5 = (change in y)/(change in x)
so change in y = -2 and change in x = 5
so the new point is (2+5,-3-2) = (7,-5)
If you need to find the slope between two points on a line:
If you are trying to find the slope of a line that goes between the points (6,2) and (3,7), you would use the slope formula:
change in y y2 - y1 7 - 2 5
slope = ------------ = ------- = ----- = --- = -5/3
change in x x2 - x1 3 - 6 -3
Notice, that you could also find it like this:
change in y y2 - y1 2 - 7 -5
slope = ------------ = ------- = ----- = --- = -5/3
change in x x2 - x1 6 - 3 3
It doesn't matter which point you start with in the slope, but it is very important that, once you start with one point for y2, you must start with the same point for x2. The most common reason for mistakes in this is when people aren't consistent; they start with one point up top, but start with a different point down bottom. This gives the wrong sign.
If you have a graph of a line you are trying to find the slope of, you must pick any two points on the line that you can find the coordinates of and then use the slope formula on those two points.
If you are trying to graph a line:
1. One way is to just plug in convenient numbers for one variable and see what the other variable equals. So, if you equation were:
A = 2L - 3
L | A
--------
0 | -3
1 | -1
2 | 1
etc. (You need at least 2 points to graph a line. Having a third point just gives you a check on your work.) You would then plot the points and connect the dots. Notice, that it doesn't matter what the variables are called; x and y or L and A (or F and C). As long as you label which axis is which on your graph, you are okay. (Generally, we let the variable the equation is solved for, A in this case, be the vertical or "y axis".) This method works okay, but doesn't give you a lot of extra useful information (see below).
2. Another method is to find the "intercepts". This means, you find where the line crosses each axis. Now, if you think about it a little, you will realize that, when a line crosses an axis (say the X axis, for example), the other coordinate must be 0. So, to find where this happens, you can just plug in 0 for each variable to find out where the graph crosses the other axis. Using the equation above:
A = 2L - 3
If you plug in a 0 for L, you get:
A = 2(0) - 3 = -3,
So the graph crosses the A axis at -3 (so at the point L=0, A=-3).
Now, plug in 0 for A:
0 = 2L - 3
Solving:
3 = 2L
3/2 = L
Or L = 3/2
So the graph crosses the L axis at 3/2 (or 1.5 if you prefer, at the point L=3/2, A=0).
We often summarize this in a table like this:
L | A
----------
0 | -3 <-- This is the "A intercept"
3/2 | 0 <-- This is the "L intercept"
If you plot these two points, you can then connect them to graph the line. This method works fine unless the line crosses through the origin (0,0), in which case your two intercepts will be at the same point, so you will only have one point to plot. (You will have to find a second point using method 1 to actually graph the line.)
3. The other method is to first put the equation in "slope-intercept form", in other words:
y = m x + b
where m is the slope and b is the "y intercept". In our equation above, we have:
A = 2L - 3
If you treat A like y and L like x (just because that puts it in the right form), this gives you:
y = 2x - 3
So, the slope is 2 (or 2/1) and the y-intercept is -3. To graph this line, you would first plot the y-intercept, so put a point at -3 on the y axis, at the point (0,-3) (why?). Then, use your slope to find another point. So, starting at that point, you go up 2 (the "rise") and over to the right 1 (the "rise"). If you look at this on a graph, you will see this gives you a second point at (1,-1). You could plot a third point by starting from this new point and moving up 2 and right 1 to the point (2,1). Then you can graph. This method is a bit harder in some respects than the first two, but it is especially useful if you need to know the slope and/or the y-intercept.
I am often asked "What do you really use slope for, if you aren't graphing a line? What good is it?"
The key to understanding how you use slope is to realize that another name for slope is "rate". So, anytime you are talking about the rate at which something is happening, you are really talking about the slope of a line. Notice that you know you are talking about rate when your units are something like "miles per hour" (rate of distance with respect to time), "miles per gallon" (rate of distance covered with respect to gas used), "milligrams per hour" (rate of dosage with respect to time), "patients per day" (rate of patients seen with respect to time), etc. (Notice the "per" in the units; this is an important clue that you are talking about a rate.)
So, for example, if you were to look at the intake of a particular medicine by two different people:
Thelma takes 4 grams every 3 days, while Wilbert takes 3 grams every 2 days.
So, who has a bigger intake here? Well, Thelma's rate (or slope) is:
4 grams
------- = 4/3 grams/day = 1.33 grams per day
3 days
While Wilbert's rate is:
3 grams
------- = 3/2 grams/day = 1.5 grams per day
2 days
So, Wilbert is taking this medicine at a faster rate (i.e., a steeper slope). (I wrote this in response to someone who was going into the medical field. This same method can be used in many other fields, however.)
Now, real life is a bit more complicated than this (for example, not everything is a line; Thelma might vary her intake, perhaps). However, this gives us a useful way to compare different rates against each other.
Finding the Equation of a Line
See the note above about graphing lines.
If you are using the point-slope form of the equation for a line to find the line that passes through the two points (4,5) and (2,3):
First, find the slope:
m = (5 - 3)/(4 - 2) = 2/2 = 1
So the equation could be either:
y - 5 = 1(x - 4)
OR
y - 3 = 1(x - 2)
Either one of these would be correct. If you put these into slope-intercept form (solve for y), you can see why it doesn't matter:
y - 5 = 1(x - 4)
y - 5 = x - 4
y = x - 4 + 5
y = x + 1
Or
y - 3 = 1(x - 2)
y - 3 = x - 2
y = x - 2 + 3
y = x + 1
So, either way you do it, you really get the same equation (even though they don't look alike when you first write them down). So, you can use either point in the formula.
Another common question I get is "How do you know when to use which equation for a line?" Here is a slightly over-simplified (though very useful) answer:
If you need to graph a line, use the slope-intercept form so you can easily find the slope and y-intercept.
If you need to find the equation of a line, use the point-slope form since you only need to know the slope and the coordinates for some point the line passes through.
Solving Systems of Linear Equations
Notice, there are different methods to solve these problems. The graphical method is great when you want to really see what is going on (for example, it helps you see how many solutions you have). However, if you want an accurate solution, the graphical method is really the best method; an algebraic method is much better at this. (Especially since often in "real life" the answers aren't always nice "whole numbers". If your answer is something like x = 3.0215, y = -2.3516, you aren't likely to be very successful finding the answer using graphs.)
I am often asked "How do I know what to multiply each equation by when using the elimination method?" For example, if you are solving the system:
2x + y = 25
2x - 5y = 7
First, you have to choose which variable you want to get rid of. It doesn't really matter which you choose, but if you have one variable that has the same coefficient in both equations, that is a good one to choose (x in this case). Then, you need to get the same coefficient for that variable in BOTH equations, but with opposite signs. To do this, you find the least common multiple of the two coefficients and figure out what you have to multiply each by to get that number. You also want them to be opposite signs, so if they aren't already, you pick one of them and multiply by a negative.
So, in this problem, I would choose to eliminate the x's (since there are the same number of each). The least common multiple of 2 and 2 is 2 (this is why we chose the x). However, they aren't the same sign, so I decide I will multiply the second equation by -1 to make it negative. So:
2x + y = 25
-1*(2x - 5y = 7) (this means multiply both
sides by -1; it isn't exactly legal
but you get the idea...)
And we get:
2x + y = 25
-2x + 5y = -7
We now have the same number of x's in each equation, but opposite in sign, so we can add them together:
2x + y = 25
-2x + 5y = -7
-------------
0x + 6y = 18
6y = 18
y = 18/6 = 3
And then you plug the y back in to either equation to solve for x.
A little more complicated example:
3x - 6y = -9
4x + 9y = 22
Now in this case, neither variable is particularly easiest to choose, so we just choose one to eliminate. Let's choose the y's (they are already opposite in sign, so we don't have to worry about that):
The least common multiple of 6 and 9 is 18. Therefore, what do we need to multiply the 6 by (in the first equation) to get 18? We need to multiply that equation by 3:
3*(3x - 6y = -9)
9x - 18y = -27
What about the second equation? We need to multiply the 9 by a 2 to get 18:
2*(4x + 9y = 22)
8x + 18y = 44
Now, add the two new equations together:
9x - 18y = -27
8x + 18y = 44
--------------
17x + 0y = 17
17x = 17
x = 17/17 = 1
Then solve for y.
Another question I am often asked is how you know whether to use elimination or substitution to solve a system of equations. In my note right above this, I talked about how to know what to multiply equations by to do elimination, but I thought I would give a quick and easy answer to this here:
If ANY of the variables has a coefficient of 1, then substitution is a good choice. On the other hand, if ALL the variables have coefficients different than 1, use elimination. So for example:
x - 3y = 7
2x + 5y = 2
x has a coefficient of 1 in the first equation, so it would be easy to solve this by substitution (solve the first eq. for x).
3x - y = 2
x - 2y = 1
Both y (1st eq) and x (2nd eq) have coefficients of 1, so you could use substitution, solving for either one of those.
On the other hand,
2x - 3y = 3
4x + 2y = 1
None of the coefficients is 1, so elimination is probably a better idea here. Of course, you can really use whichever method you feel like in any case. It isn't like one is "right" and the other is "wrong". It's just that, if you follow my advice here, it is easiest. (It's perfectly fine to use elimination if one of the variables has a coefficient of 1. However, if none of them is 1, using substitution will probably wind up giving you more fractions to deal with than elimination would, which is generally something most people prefer to avoid.)
When you are factoring by grouping, you sometimes need to rearrange things. Remember, your goal is to group things in sets of 2 terms so that when you factor everything you can out of each set, you wind up with the same thing left over.
18x^2 - 10y - 15xy + 12x
So, let's arrange the terms with the x together and the ones with the y together:
18x^2 + 12x - 15xy - 10y
Now, group them together in sets of 2 terms:
18x^2 + 12x - 15xy - 10y
-1st group - -2nd group-
Now, factor common terms out of each group:
6x (3x + 2) - 5y (3x + 2)
-1st group- -2nd group-
Now, notice that each "group" has a (3x + 2) in it, so that means you can factor that out of each one:
(3x + 2) ( 6x - 5y )
-1st group- -2nd group-
= (3x + 2) (6x - 5y)
IMPORTANT: Remember that this method (factoring by grouping) doesn't work unless both groups have the same factor in them. So, in this case, they both had 3x + 2 in them. If it wasn't the same, then you have to try again (so, perhaps rearrange the terms or look for some other factoring method).
Factoring polynomials - How do I know which method to use?
Okay, so you have now (hopefully) learned all the different factoring methods. A pretty common question I get asked at this point is "How do you know which method to use?" There is good news and bad news: The bad news is that there is no rule that will always tell you which method to use. HOWEVER, the good news is there are some hints that can help you out. When you have something to factor, you should always start by factoring out any common factors. This is very important. If you don't do this first, it can make it much harder to factor. After that, the simplest way to pick a method is to count the number of terms. Based on this, it can give you a clue what method to try. So, if you have:
Solving Quadratic Equations by Factoring - A Warning
A kindly warning: After people learn how to solve quadratic equations by factoring, they seem to like it so much they can't stop. So I present a cautionary tale of two similar but different problems:
Notice how the answer is different: The first problem asks you to FACTOR and your answer is just an expression factored:
(x - 2)(x - 4).
The second problem asks you to SOLVE and your answer is:
x = 2, x = 4 (i.e., numbers).
The problems are very similar, but it is important that you are very clear on which one is which.
Solving qudratic equations with the quadratic formula (or not)
You will not have to memorize the Quadratic Formula for this class (but you will for later classes, so you might want to do so anyway). Just a reminder: If you are asked to solve a quadratic equation and you can't factor it, you cannot say "no solution" or "not factorable". This is when you need to use the quadratic formula.
On a slightly different topic, I often get asked the following question: How do I know when to factor and when to use the quadratic formula to solve an equation? Well, it's like this:
-(-1) +- Square root ( (-1)^2 - 4 (1) (-1) )
x = --------------------------------------------
2 (1)
1 +- Square root ( 1 + 4 )
x = ----------------------------
2
x = (1 + Square root (5) ) / 2 = 1.62 and