This month has been unexpectedly crazy. I apologize for the lack of any real content here.
I have another link to share with you, but you can look forward to more authentic Dead Bunny goodness next week!
Today’s find shows how to find the equation for a line. It covers everything from some of the equations for a line, to how to derive a line’s equation from given information, to find a line parallel to a line through a given point. It’s a lot of useful information in a compact space.
Real posts are coming, never fear! (Dead Bunny has had quite the week!)
Until then, consider the power of learning your basic math facts:
Memorize Those Facts!
There’s enough research out there to indicate that children who commit the multiplication facts to memory have greater academic success in mathematics. Yes, the chronic complaint heard from teachers over and over again is that their students just don’t know their facts. I can’t emphasize enough how important it is to commit those facts to memory. With a already packed math curriculum, it’s important that children get support at home to learn the facts. (Source)
Mastering math is sometimes all about the attitude.
I’m constantly telling my students this. If you tell yourself you can’t do math, then you’ve already lost. The minute you change your mind and tell yourself you can do math, then you increase the chances that you’ll learn it.
You may not pick up a concept on the first attempt, and that’s okay. As long as you keep working at it, you will get it (especially if you are able to explore the concept different ways until you find the one that makes sense to you).
Incidentally, this is true of most things in life.
(Sorry about this week, everyone. The past two weeks have been fairly crazy both at work and in my personal life, so I haven’t had the time to write the next series of articles for you guys. I’m planning on getting those done this weekend!)
So far, we’ve solved a system of equations by graphing and by substitution. Today, we tackle the challenging method of solving a system of equations by elimination.
You’ve probably already figured out that something is going to get eliminated, and you’re right. When we solve by elimination, we eliminate one variable in order to solve for the other.
Let’s consider the following system of equations.
x + 3y = 6
2x – 3y = 3
The first thing we’re going to do is add these two lines together.
x + 2x = 3x
3y + (-3y) = 0
6 + 3 = 9
This leaves us with the new equation: 3x = 9. Notice that the y terms cancelled each other out. This is the elimination this method is named for.
When we solve 3x = 9, we find that x = 3. Now we plug 3 in for x in either equation. I’m going to use the first equation for this example.
3 + 3y = 6
Now I solve for y as usual.
3y = 3
y = 1
The solution, the one point that lies on both lines, is (3, 1).
Generally, when you look at a system of equations, one of the methods will jump out at you as the easiest method to use for the problems. For systems where you have an x or a y with no coefficient (the number in front of the variable), you’re probably best off using substitution. If you have the positive of a term, and its exact negative in the other equation, go for elimination.
Today is 3/14, the day some geeks like to call “Pi Day” because Pi’s numerical value starts with 3.14…
So, what is Pi? Pi, represented by the Greek letter π, is the ratio of a circle’s circumference to its diameter.
Try out this little experiment:
You may notice as you continue that the numbers are all right around 3.14. Despite the fact the ancient Babylonians and Egyptians were aware of approximations of Pi., the ancient Greek mathematician Archimedes was the first to determine the actual value of Pi as a non-terminal, non-repeating decimal that begins 3.141592.
Go forth! Celebrate Pi Day (perhaps with a bit of pie?)!
The other day, we talked about solving a system of equations by graphing. Today, we’re going to look at solving a system of equations by substituting.
As the name suggests, this method involves substituting. This involves solving one equation for either x or y, and then placing that equation into the other to solve for the other variable. This is a bit labor-intensive, but it’s worth it since many find this easier than using elimination to solve a system.
Let’s solve this system of equations by substitution:
x + y = 12
2x + 3y = 9
We’re going to start by solving the first equation for y. We do this by subtracting x from both sides of the equation, leaving us with y = 12 – x.
Now, we’re going to substitute 12 – x everywhere we see y in the second equation. We can then solve for x. The whole process looks like this.
2x + 3(12 – x) = 9
2x + 36 – 3x = 9
2x – 3x + 36 = 9
-x + 36 = 9
-x = -25
x = 25
We now know the value for x, so we can substitute that in for the x in either equation. For this example, I’m going to substitute it back into the first equations.
25 + y = 12
y = -13
The one point that lies on both lines, therefore, is (25, -13).
We have one more method, elimination. Look for that one later this week!
Sometimes, you need to know what points on a graph make two or more equations true. This is called solving a system of equations. There are a few ways you can do this.
The first is to graph the lines and see where they meet.
Let’s look at the graph of y= -3 and y=x. We graph both on the same graph and get this:
We look at the graph to visually tell where they intersect. In this case, they meet up right around (-3, -3). So the solution for this set of equations is (-3, -3) because it’s the only point that is on both lines.
Some students find this the simplest method for solving a system of equations. Others find this method a good way to double check their answer when they solve a system using a different method.
The other methods, solving by substitution or by elimination, will be covered later this week.
I’ve actually blogged about this before, but I think it’s both interesting and useful to see it all in one place.
Prime factorization is probably the quickest, most foolproof way to determine the GCF or LCM for two numbers. In fact, prime factorization gives you everything you need to find both the GCF and the LCM in a single bit of work.
Let’s say we’re asked to find the GCF for 12 and 16. We’ll start by finding their prime factors.
12: 2 * 2 * 3
16: 2 * 2 * 2 * 2
To find the GCF, we look for common prime factors.
12: 2 * 2 * 3
16: 2 * 2 * 2 * 2
Both lines have 2 *2 bolded, so I multiply them together and learn that the GCF for 12 and 16 is 4.
But what if we’re suddenly asked to find the LCM of both numbers instead. Fortunately, we just found the prime factors for 12 and 16 for our GCF problem. Let’s look at them now to see how we can use them to solve for LCM.
12: 2 * 2 * 3
16: 2 * 2 * 2 * 2
We have the exact same prime factors, but how can we use that to find the LCM? Oddly enough, we do it by using the numbers left over after we found the GCF.
12: 2 * 2 * 3
16: 2 * 2 * 2 * 2
For 12, we were left with 3; and for 16, we were left with 2 * 2, or 4. That’s pretty interesting.
We now multiply 12 and 4 together to get 48, the LCM for 12 and 16. Just to be sure, though, we also multiply 16 and 3, and discover that it also equals 48.
If we have the prime factors for two numbers, we use the ones they have in common to find their GCF and the ones they don’t have in common to find LCM. We’ve just used the same information to cut our work in half.
If you square the square root of a number, you end up with the number itself.
Hmm…that sounds a little confusing. Let’s see this one in action.
We’re going to begin with √9.
Now let’s square it. (√9)2. This is the same as saying: √9 * √9
Because we happen to know our perfect squares (You do know your perfect squares through 202, right?), we know that √9 = 3.
We plug the 3 back into our equation every time we see a √9, and we get 3 * 3, which equals 9.
For those of you familiar with radical expressions, let’s try this one more time. √9 is the same as 9½.
When we square it this time, it looks like this: (9½)2.
We have a law of exponents that says when we raise a power to another power, we simply multiply the two together to get the new exponent. In this case, we’re multiplying 2 and ½ which equals 1. Our problem now looks like this: 91.
Any number raised to the power of 1 is itself, so our answer is 9.
I’ve talked about finding the LCM by exploring the multiples of both numbers. Earlier this week, I was asked to figure out how to determine the least common multiple using the prime factorization of both numbers. (It came up on a student’s homework.)
This one is kind of neat, and actually takes less time than the traditional method. Let’s take a look.
We’re going to find the LCM for 12 and 16. So we’ll start by finding the prime factors for both numbers.
12 factors to 2 * 2 * 3
16 factors to 2 * 2 * 2 * 2
Both 12 and 16 have two 2s in their list of factors, so we’ll ignore those. That leaves us with the following factors.
12: 3
16: 2 * 2
To find the least common multiple, we multiply the original number by the remaining factors of the other number.
12 * 2 * 2
16 * 3
If you multiply both lines out, you’ll find they both equal 48. The least common multiple for12 and 16 is 48.
You’ll notice this took a lot less work than the other LCM method, so you might want to try it out yourself the next time you’re staring down a page full of LCM questions.
