READ ME!!! <3

[Sweet Ichigo]

Okay...I hope I'm putting this topic at the right spot. *stares* Hmm...Oh well, Anyhow, I was wondering if anyone in here is great at math. Like are ya confindent w/ your math skills.

http://i3.photobucket.com/albums/y63/SasukeLUV/math.jpg <--Something like this?

[Sepho]

My algebra teacher was fond of saying "The problem with word problems is, there's too many words". I'm not sure how this piece of advice is applicable to the situation at hand, but the man was brilliant, and his genius deserves to be exposed to the world.

Alebgra needs to be picked apart and completed methodically to find the logic behind some of the problems. Figuring out what they're <i>asking for</i> is the biggest part of the problem, I think. For example, problem 2-2 reads (and I'm going to try to be is descriptive as possible):

<i><b>What is the only pair of integers (x,y) for which twice the square of the first equals three times the square of the second</b></i>

Picking it apart will tell you that you're trying to find the value of two variables. In addition, the sentence more or less gives you the structure of the equation.

<i>What is the only pair of integers</i>...

You have two variables (<i>pair</i>)

...<i>for which twice the square of the first</i>[variable] ...

<font color="#FFFFAA">2x²</font>

...<i>equals</i>...

This literally translates into an equal sign (<font color="#FFFFAA">=</font>)

...<i>three times the square of the second</i> [variable] ...

<font color="#FFFFAA">3y²</font>

You'll end up with the following equation:

<font color="#FFFFAA">2x² = 3y²</font>

And you can solve from there.

In problem 2-4, the wording can be tricky if you haven't seen it before.

<i>The squares of two consecutive positive integers</i>.

When you're dealing with a problem where it asks for consecutive integers, (i.e. 1 and 2, or 778 and 779, for example), these are represented in the equation by <font color="#FFFFAA">x</font> and <font color="#FFFFAA">x+1</font> (do you see the logic behind this?). If the problem specifies consecutive odd or even integers (i.e. 2 and 4, or 17 and 19, for example), these are representated in the equation by <font color="#FFFFAA">x</font> and <font color="#FFFFAA">x+2</font>. Got it?

The problem reads:

<i><b>The square of two consecutive positive integers differ by 1987. What is the sum of these two integers?</b></i>

You know you have two "variables" (okay, so it's one variable, but that's not important) in the equation, (<font color="#FFFFAA">x, x+1</font>), because of the <i>two consecutive positive integers</i> part. Keep in mind, though, that the problem asks for the squares of these variables. Thus, you'll be working with <font color="#FFFFAA">x²</font> and <font color="#FFFFAA">(x+1)²</font> which FOILs (do you know how to FOIL?) out to <font color="#FFFFAA">x² + 2x + 1</font>.

Now, let's look at another part of the problem:

...<i>differ by 1987</i>...

The "differ" part implies that there's a difference between the two values, and that difference is 1987. Because of that, it's logical to think that if you subtract one value (the lesser value) from the other (the greater value), you'll get 1987 as an answer. Since <font color="#FFFFAA">x+1²</font> is no doubt the greater value, we'll subtract <font color="#FFFFAA">x²</font> from it. Let's set it up:

<font color="#FFFFAA">(x+1)² - x² = 1987</font>

From there, we can FOIL (like I showed you above), and we get:

<font color="#FFFFAA">(x² + 2x + 1) - x² = 1987</font>

*Note that <font color="#FFFFAA">x² + 2x + 1</font> is just the FOILed version of <font color="#FFFFAA">(x+1)²</font>. They're completely equal in value.

From there, we'll just simplify. The positive and negative <font color="#FFFFAA">x²</font>'s cancel out, and you'll get:

<font color="#FFFFAA">2x + 1 = 1987</font>

Then you just subtract 1 from each side, to leave <font color="#FFFFAA">2x</font> by itself on the left side of the equal side. You'll get:

<font color="#FFFFAA">2x = 1986</font>

After that, all you have to do is divide everything by two to get <font color="#FFFFAA">x</font> by itself. So, the value of <font color="#FFFFAA">x</font> is...

<font color="#FFFFAA">x = 993</font>

And <font color="#FFFFAA">x+1</font> is <font color="#FFFFAA">994</font>, so your two consecutive integers are <font color="#FFFFAA">993</font> and <font color="#FFFFAA">994</font>. Let's go back and check to make sure.

<font color="#FFFFAA">993² = 986,049
994² = 988,036

988,036 - 986,049 = 1987</font>

It all checks out.

I'mma stop there for now and wait for a reply.

[Odaisé Gaelach]

Brilliant work Sepho. Excellent!

I'm going to try and figure out 2-5, 'cause I know it's got something to do with intergration...

[Sepho]

Thank you

Try 2-4 while you're at it. I ran out of time earlier, but I'm going back to 2-4 right now and I can't solve it.

I end up with:

EDIT: Now that that issue is taken care of, I'll post how to setup the problem back in the first post.

[Kirkpatrick]

Sepho, I think you got caught by your own explanation - it asks for two consecutive positive integers, so it would be x, x+1 rather than x, x+2, which works out to a whole number.

[Sepho]

But, but, but...

If the integers are both positive (or odd), they would have to increment by 2 (x, and x+2). If they incremented by 1, then one integer would be odd, and the other would be even.

[Kirkpatrick]

But, it only requires that they be positive (i.e. x > 0), not that they both be even or odd. It specifies that they be consectutive, so they would only be one whole number apart.

[Sepho]

Okay, okay, okay.

I for some reason keyed in on the positive part and my mind started to play tricks on me. Okay then. The problem is easy enough once that's out of the way.

I thought I was going insane. G'job.

[Sweet Ichigo]

O_O *in shock & already insane* What?!! Oh,......I got it. >.> I think. Yea....Anyhow, thanks a lot you two. >( =o])

++ EDIT: Haha. I found all the answers to the problems. And it's 100% I did a research online and found it..


http://www.mathleague.com/hstest/87-...-solutions.htm

So proud of myself.