PDA

View Full Version : Someone who knows some math please help me!


Jason1
02-25-2002, 03:56 PM
First off, lets get one thing straight. Im in a math calss that is way too hard for me. I shouldnt be in it, but I tested into it, so now im stuck in it. They are making us give an oral report of sorts on a math problem.(you would think they would cut us breakes from this ****, considering im in a smarter class, but no, we get to do fun projects like this ****.) So, normaly I would try to do this myself, but see, this is the problem. We sign up for a problem that was on our homework, and then the day we grade that homework assignment, is the day we do our reports. So my teacher wont explain the assignment(and how to do my problem) until the day before Im doing my report. And im going to be busy the night before my report, so wont have time to work on it then. Most people in my class allready know how to do these problems, so they dont have that problem, but im the dumbest one in my class, struggling to maintain a C- Average. So we have to explain the problem on the blackboard. So im going to give the problem I singed up for, and I want someone to plese explain how to get the answer, explaining the steps you took to get there. Im doing my report on Thursday. Here is the problem.

The Directions State:

Solve by either the substitution method or the addition-or-subtraction method.

My problem is:

x-5y=2
2x+y=4

the directions at the begining of the lesson tell you how to do it using the Addition method...this is how it says...

1. Add similar terms of the two equations.
2.Solve the resulting Equation
3. Substitute the number you get for x in either of the origional equations to find y.
4. Check in BOTH equations.

so you should end up with an ordered pair. I appreciate any help, although I probably wont get any. The most crucial thing is getting the correct answer, because if you dont, you get a zero.

Neo
02-25-2002, 04:04 PM
Either multiply the bottom equation by 5 or the top equation by -2. Then add them.

azazela31
02-25-2002, 04:16 PM
Patheticly easily!!

x-5y=2
2x+y=4

x=2+5y

2(2+5y)+y=4
4+10y+y=4
11y=0
y=0

to find X

2x+y=4

substitute in for y

2x+0=4
2x=4
x=2

check for both equations

x-5y=2

2-5(0)=2
2=2 (first equation is true)


2x+y=4

2(2)+0=4
4=4

last equation is true



Now you better kiss my ass for waisting my time doing this.

Xantar
02-25-2002, 04:54 PM
Well, hopefully, Neo and azazela were helpful. If not, I'll explain this to you in as much excruciating detail as you can stand.

First off, I hope you understand algebraic addition, multiplication and a little bit on manipulating equations. Otherwise, you are truly screwed.

Okay, so your two equations are

x - 5y = 2
2x + y = 4

I am now going to explain what your directions mean.

1. Add similar terms of the two equations.

A term is a variable (a letter) multiplied by a constant (a number). x and y are considered terms because they are actually multiplied by 1. Some people consider numbers by themselves to be terms, but don't worry too much about it.

Similar terms are terms with the same variable. So x and 2x are similar terms. -5y and y are terms. Note that I put the negative sign in front of -5y. In algebra, subtraction is considered to be the addition of a negative number rather than an operation all unto itself.

So to add the similar terms of the two equations, add the two terms with x together, add the two terms with y together and add the two numbers together. x + 2x = 3x, -5y + y = -4y, 2 + 4 = 6. Since all the variables are on the left of the equation and the constants are on the right of the equation, you can just stick all the variables in the same place. So using the results from that addition I just did, you'd get the equation 3x - 4y = 6.

2.Solve the resulting Equation

What they mean is "solve the resulting equation for x" since the next instruction requires you to substitute a number for x. Unfortunately, this leads to some truly ugly fractions that I don't think you want to deal with (you'd get x = 2 + 4/3y). This is the other trick with solving by addition. You can multiply an entire equation by a constant. This is what Neo was talking about in his post. So let's say you choose to multiply the bottom equation by 5. Your system of equations is now

x - 5y = 2
10x + 5y = 20

Do you see how if you add those two equations, you'd eliminate y altogether? In other words, adding them gets you the equation

11x + 0y = 22

Which is the same thing as

11x = 22

I'm pretty certain you can figure out from here that x = 2 (if you're at least maintaining a C-, you can't be stupid enough to not understand how that works).

3. Substitute the number you get for x in either of the origional equations to find y.

Ok, you just got a number for x. Namely, x = 2. So pick one of the equations and write it down, but every time you see an x, just write the number 2 instead. Remember, 2x means 2 multiplied by x. So if x = 2, then 2x is 2 times 2, not 22. I didn't think you would actually write 22, but I told you that I'd explain everything in excruciating detail.

Anyways, pick one of the equations and stick in 2 for x. Let's say you pick the top equation. The equation now becomes

2 - 5y = 2

You should be able to solve this, and for once, I'm not going to tell you what the answer is (even though azazela already did).

4. Check in BOTH equations.

So that means whatever value you got for x, stick it into both equations, and whatever value you get for y, stick it into both equations. Then add everything up. If the numbers on both sides of the = sign are equal, then you've got the correct answer.

I can show you solving by substitution (which is a far superior method, in my opinion), but the directions didn't require you to know how to do that.

By the way, couldn't one of your parents have explained this to you?

*looks back over post*

Man, I have way too much time...

Ric
02-25-2002, 04:56 PM
This is why I got a D for Maths and dont plan on retaking.

Kitana85
02-25-2002, 04:56 PM
Well, there's nothing to add, except, yep, they both look right.

azazela31
02-25-2002, 05:04 PM
If you don't understand it just copy down what I wrote and come up to the teacher for exta help.

Yoda9864
02-25-2002, 05:15 PM
Wow, I remember those easy days of simple algebra (no offense Jason1). Just wait til calculus, which in it's self isn't too incredibly hard.

BreakABone
02-25-2002, 05:22 PM
Originally posted by Yoda9864
Wow, I remember those easy days of simple algebra (no offense Jason1). Just wait til calculus, which in it's self isn't too incredibly hard.

You think Cal is easy..You think Cal is EASY!!!!.. ARGH

This problem was nerve wrecking and annoying...

A tangent line drawn to the parabola y=4-x² at the pont (1,3) forms a right triangle with the coordinate axes. The area of the triangle is....

gekko
02-25-2002, 05:30 PM
Here's how I do it.

Step 1: Buy a TI-89 calculator
Step 2: Enter: solve(x-5y=2 and 2x+y=4,{x,y})
Step 3: Write down what it tells you: x=2 and y=0

:)

Jason1
02-25-2002, 05:56 PM
I understand everything just fine, except for one part which is just something im doing wrong and I shouldt be.

In Xantars Explination, he gets 2-5y=2.

now its obvious y=0, but for some reason when I solve that the proper way, I dont come up with that. So sombody please explain how to get y=0 for that....

nWoCHRISnWo
02-25-2002, 06:19 PM
Because if y is equal to zero, then it'd be 2-(5 x 0)=2. 5 x 0= 0. So you check, and 2-0=2, so it's correct.

But you said it was obviously zero, so you probably already knew that. :D So I don't really know what you're asking now...

Xantar
02-25-2002, 06:22 PM
Jason1:

I can't help you unless you post the steps you took and the numbers you got. Otherwise, all I can say is "You did it wrong. Try it again."

Revival
02-25-2002, 06:34 PM
I remember what I had a hard time with this.. it was only a month ago.. :D

Jason1
02-25-2002, 07:07 PM
eh, who cares, Ill just choose the other problem, and then everything is all okely dokely!

And Xantar, you asked why my parents cant help me, well thats because my mom was horrible at math as a Kid, and she still is today, she has trouble doing my brothers 5th grade math.

My dad was fairly smart, but hes forgotten everything, he cant help me either(except on some rare occasions)

DeathsHand
02-25-2002, 07:11 PM
Originally posted by gekko
Here's how I do it.

Step 1: Buy a TI-89 calculator
Step 2: Enter: solve(x-5y=2 and 2x+y=4,{x,y})
Step 3: Write down what it tells you: x=2 and y=0

:)

Now that sounds like my kinda mathish stuff!

Or writing down "I dunno, you tell me."

.... :D

Jason1
02-27-2002, 03:51 PM
Hey, Xantar, could you please solve the same problem for me using the substitution method. You dont have to be quite as descriptive, but make sure I still know what your talking about. Thank you.

Xantar
02-27-2002, 04:27 PM
Well, I just turned in my psychology paper, so I have nothing better to do.

*hears voices telling him to finish his fanfic*

I was going to do that next!

Anyway, here are your equations again.

x - 5y = 2
2x + y = 4

Substitution means you take one of the equations and get one of the variables all by itself on one side. Hopefully, you know how to move terms from one side of the equation to the other. Just think about it this way: when given a true equation (i.e. 1 + 1 = 2), you can add the same thing to both sides of the equation and it would still be true (1 + 1 + 2 = 2 + 2). You can also multiply both sides by the same number, and it would be true. So in any case, let's say you pick the first equation. If you add 5y to both sides, you'll get

x = 2 + 5y

This is also a true equation. Now that we know this, you can write "2 + 5y" every time you see "x" in the other equation. So the second equation will look like this

2(2 + 5y) + y = 4

Using multiplication (you do know how to multiply in parentheses, don't you?), you can get

4 + 10y + y = 4

which leads to

4 + 11y = 4

If you subtract 4 from both sides, you get

11y = 0

Obviously, then, y = 0. Going back to the first equation, you can plug in 0 every time you see "y" and figure out what x is. In this case,

x - 5(0) = 2

which means x = 2.

You can also start out by solving for y. You can solve for y in the first equation, but that gets ugly. Besides, it's not necessary because in the second equation, you can just move the x term to the other side of the equals sign and you'll have a very neat y = 4 - 2x. You can go from there, plugging in "4 - 2x" into the first equation wherever you see "y" and solve it from there in pretty much the same way as I did before. No matter how you start it out, if you do your math correctly, you will always get the same answer for x and y.

Jason1
02-27-2002, 04:46 PM
Thanks Xanny, your the man.

designated_GM3R
02-27-2002, 05:22 PM
I thought Substitution was a fairly easy lesson.
Graphing on the other hand sucks balls.

BreakABone
02-27-2002, 05:48 PM
*Realizes no one helped him with his math problem*

Thank you all..*CRIES*

Revival
02-27-2002, 06:59 PM
Originally posted by BreakABone
*Realizes no one helped him with his math problem*

Thank you all..*CRIES*
*points and laughs*
What math problem?

Xantar
02-27-2002, 09:19 PM
Originally posted by BreakABone

A tangent line drawn to the parabola y=4-x2 at the pont (1,3) forms a right triangle with the coordinate axes. The area of the triangle is....

Edit: stupid signs switching on me...ok, here's my second try.

If you take the derivative of the parabola's function, you get dy/dx = -2x. You know that the line intersects at the point (1,3). In other words, where x = 1. Plugging that into the derivative function of the parabola, you get dy/dx = -2.

So now you know the slope of the line is -2 and it passes through the point (1,3). You know how to come up with the equation fo the line from here (I presume). If you want to do this on your own from here, don't read any further. Just check back here to verify your answer.

Are you gone yet? Ok, here goes.

The equation you should get is y = -2x + 5. So you know that the x-intercept is 5/2 and the y-intercept is 5. If you draw everything out on graph paper, you'll see that the base and height of the triangle are 1/2 unit long and 1 unit long respectively. So using the area of a triangle formula, you find that the area of the triangle is 5/2 x 5 x 1/2 or 25/4.

BreakABone
02-27-2002, 09:37 PM
Originally posted by Xantar


If you take the derivative of the parabola's function, you get dy/dx = 2x. You know that the line intersects at the point (1,3). In other words, where x = 1. Plugging that into the derivative function of the parabola, you get dy/dx = 2.

So now you know the slope of the line is 2 and it passes through the point (1,3). You know how to come up with the equation fo the line from here (I presume). If you want to do this on your own from here, don't read any further. Just check back here to verify your answer.

Are you gone yet? Ok, here goes.

The equation you should get is y = 2x + 1. So you know that the x-intercept is -1/2 and the y-intercept is 1. If you draw everything out on graph paper, you'll see that the base and height of the triangle are 1/2 unit long and 1 unit long respectively. So using the area of a triangle formula, you find that the area of the triangle is 1/2 x 1/2 x 1 or 1/4.

Well the derivative is wrong..which would throw off your whole problem if done right

But the question is done incorrectly.. I myself can't do it.. but I know the answers and the steps.. just hoping someone could help explain it to me

Revival
02-28-2002, 09:36 AM
*explodes*

Stupid math.. :mad2:

fingersman
02-28-2002, 10:31 AM
Originally posted by BreakABone


Well the derivative is wrong..which would throw off your whole problem if done right

But the question is done incorrectly.. I myself can't do it.. but I know the answers and the steps.. just hoping someone could help explain it to me


The derivative is right as far as I can tell. x squared gives you 2x. Oh well maybe I read it wrong or something. :D

BTW how old are you Breakabone?

Revival
02-28-2002, 10:58 AM
Mr. Earl Rufus is currently 16, and will be turning 17 on April 2. :D

El Gaucho Goof
02-28-2002, 03:57 PM
Well the derivate of x squared (Don't know how to do it on a Mac) is 2x.. but you also ned to take into account the minus sign in front of it.. Also the problem Xanny did is different from the one I have quoted

Originally posted by Shooter
Mr. Earl Rufus is currently 16, and will be turning 17 on April 2. :D

As for shooter, he really scares me sometimes, but that is all correct.

Neo
02-28-2002, 04:23 PM
sdfsdf*test*

Revival
02-28-2002, 04:26 PM
Originally posted by El Gaucho Goof
As for shooter, he really scares me sometimes, but that is all correct.
Heh.. your birthdate is in your profile, and since I hate math, I suck at it, I then went to April on the calendar to see how old you were going to be on April 2nd, and it said 17, so you had to 16 now. :D

Simple eh?