Excuse me, could you help with my maths homework?

[hack124x768]

Ok, So I have an algebra project due tomorrow and I'm completely bamboozled on how to do something. I need to show graphically and algebraically where a meteor and the earth will collide, assuming both are on a plane with the center of earth as the origin. I have the answers (Back of book), But I havn't been able to get to them, so they do me no good. Sorry, I don't know how to make math symbols here, So I used x(sqr) to mean x squared and sqrt(x) to mean square root of x.

The meteor's motion is modeled by:
x(sqr)-4y(sqr)+80y=340

And earth is a circle made by:
x(sqr)+y(sqr)=40

So from what I gather I need to solve the meteor's equation for y so I can graph it as a function, (earth is no problem y=sqrt(40-x(sqr)) ) I think I need to use the quadratic function to do this, But I havn't had much luck. I got -397.49 and 237.49.

For the algebraically part, I just need to solve it with a system of equations, but I'm not having much luck there either.

I've asked my teacher for help with this several times, and He can't think of a way he can help without just giving me the answers (which I have anyway, so who cares?), So I was hoping someone here could explain it better.

Thanks.

Subject edited because it was retarted
Hey Mod! You left an s on math. --hack
That's because it's mathematics and because I'm not American.
Eh, whatever. Just wait till I get my mod status, I shall have revenge!

[SpitValve]

Typically we use ^ for exponents, so x^2 is x squared.

You have

x^2 - 4y^2 + 80y = 340 (1)
x^2 + y^2 = 40 (2)

(1) can be rearranged as x^2 = 340 - 80y + 4y^2.
Substitute this into (2). Then solve for y and substitute back into (1) or (2) to get x.

That's as much hint I'm going to give you, because this is homework...

[hack124x768]

Typically we use ^ for exponents, so x^2 is x squared.

You have

x^2 - 4y^2 + 80y = 340 (1)
x^2 + y^2 = 40 (2)

(1) can be rearranged as x^2 = 340 - 80y + 4y^2.
Substitute this into (2). Then solve for y and substitute back into (1) or (2) to get x.

That's as much hint I'm going to give you, because this is homework...

(Sorry, been a while since i've done math on a computer and I forgot about the ^ symbol)
By substitute, do you mean to subtract (1) from (2)? Or replace x (or y?) with (1)? My current math book doesn't cover it, as we did that last year (should have done the homework) and havn't used it since, so does anyone know of a good way to explain it or a site that can?

[TheTankengine]

solve one for x, then plug that into the second equation. Then you will be able to solve for y. Once you know y, you can plug that into either eqn and get x.

Yay!

[SpitValve]

ok... put it this way

If you have y=x^2+z^2 (1)

and x+y=z (2)

You substitute by replacing the y in (2) with the y in (1), so it becomes

x+(x^2 + z^2)=z.

And if you do it in your case you should get your solutions.

Alternately, you could just subtract your (2) from (1), which would just give you the same equation in the end anyway.

[hack124x768]

ok... put it this way

If you have y=x^2+z^2 (1)

and x+y=z (2)

You substitute by replacing the y in (2) with the y in (1), so it becomes

x+(x^2 + z^2)=z.

And if you do it in your case you should get your solutions.

Alternately, you could just subtract your (2) from (1), which would just give you the same equation in the end anyway.

So, the (2) would turn into
(340 - 80y + 4y^2) + y^2 = 40
Simplified
5y^2 + 80y = -300
right?

So how would I solve that for y? Would I use 5, 80, and +300 for a, b, and c values in the quadratic equation?

[TheTankengine]

Go forth, little grasshopper. The world is large.

[hack124x768]

Go forth, little grasshopper. The world is large.

Does that mean google it? Or go for it? I'm gunna say go for it...


Ok, So (-80 +/- sqrt(80^2 - 4 * 5 * 300) / 2 * 5) gives me -30 and -130.
What do those mean? Are they horizontal and vertical dilations? Or are they y values I should try using?

[hermaj]

If you're having trouble subbing (1) into (2), is it easier if you try and sub (2) into (1)? That's what I did.

40 - y^2 - 4y^2 + 80y = 340

Does that maybe look a little bit better? I don't know.

EDIT: Also, do you know how to solve quadratics? You seem a little heavy on the square rooting there, but really you need to do very little with square roots. Quadratic equations were the only thing I knew in maths; probably because I was away every time we covered them each year and had to have it explained extra times

[hack124x768]

Ok, What about the graphically part? There should be an easy way to convert x^2 - 4y^2 + 80y = 340 to a y= function. I tried just switching x and y and then solving for y, thinking it would give be the same thing, but sideways, and that didn't work out. If it helps at all I have a Texas Instruments TI-84 calculator to graph it on.

[hermaj]

Well I'd say you need to know the answers to graph it. You have the answers, right, from Bob?* If you have the answers it should be pretty obvious what kind of graph you need to do, I think. And if you have a graphing calculator doesn't it do the thinking for you? I don't know, I've never had one.

But really, if I were you I'd just work on solving the thing. *shrugs* It's not too hard - sub one equation into the other, bring everything over to one side, bring everything down to the lowest common denominator, solve the quadratic, sub the values back in. Make sure you don't get your signs muddled up along the way. I think that is all the help I should be giving you. But I remember what it was like to hit a brick wall in maths and it was really frustrating.



* I don't really know if this was in any other school, or really anywhere other than my group, but whenever we got an answer from the back of the book, we were like "Oh, yeah, Bob helped me out with that one." We were lamez0rz, really.

[TheTankengine]

Go forth, little grasshopper. The world is large.

Does that mean google it? Or go for it? I'm gunna say go for it...


Ok, So (-80 +/- sqrt(80^2 - 4 * 5 * 300) / 2 * 5) gives me -30 and -130.
What do those mean? Are they horizontal and vertical dilations? Or are they y values I should try using?

It meant go for it, in the most cryptic way possible, I guess.

You solved for y, so those are your y values (yes, complicated I know). Plug those into an equation and you will get the respective x values.

You will not be able to plot ax^2 + by^2 = c, because it is not a function (does not pass VLT). Basically, you just split it up into two functions (hint:split it horizontally).

[hack124x768]

Well, Bob says that the earth and meteor intersect at (-2,6) and (2,6).
I don't think -30 and -130 work for that pair. Where did I go wrong?

(May as well say it now, Thanks guys.)

[edit[Tried using 130 and 30 for x and y dilations in a hyperbola, no luck[/edit]

[hermaj]

So (-80 +/- sqrt(80^2 - 4 * 5 * 300) / 2 * 5) gives me -30 and -130.
What do those mean? Are they horizontal and vertical dilations? Or are they y values I should try using?

I would say here is a good example of where you could have gone wrong. Just ignore your ability to square root stuff; I know there are squares going on but it's not actually very helpful with this sort of equation. Maybe try being patient... start off with

40 - y^2 - 4y^2 + 80y = 340

Simplify that as much as possible and bring it all to one side, so that your simplified equation is equal to 0. No calculating! Just try and simplify. You should end up with something that looks like ay^2 + by + c = 0, I hope.

[hack124x768]

So (-80 +/- sqrt(80^2 - 4 * 5 * 300) / 2 * 5) gives me -30 and -130.
What do those mean? Are they horizontal and vertical dilations? Or are they y values I should try using?

I would say here is a good example of where you could have gone wrong. Just ignore your ability to square root stuff; I know there are squares going on but it's not actually very helpful with this sort of equation. Maybe try being patient... start off with

40 - y^2 - 4y^2 + 80y = 340

Simplify that as much as possible and bring it all to one side, so that your simplified equation is equal to 0. No calculating! Just try and simplify. You should end up with something that looks like ay^2 + by + c = 0, I hope.

-3y^2 + 80y - 300 = 0

Aaahhhh, So my A value was wrong... Well, that didn't seem to help much, Now my quadratic function is outputting -330 and 170, Or did you mean that when you said no calculating? Sorry, I'm just trying to be as little of an answer mooch as possible, it doesn't seem to be working well.

[aldimond]

So, the (2) would turn into
(340 - 80y + 4y^2) + y^2 = 40
Simplified
5y^2 + 80y = -300
right?

Your mistake is a sign error in this step.

[hermaj]

a) -y... - 4y... = -5, not -3.

Once you have the -5 out there you'll see all the numbers there are divisible by -5, yeah? Make sure you don't mess up the signs when you divide by -5. When you divide by a negative number, all the signs change. So, for example, 8y - 16x would become -4(-2y+4x).

[hack124x768]

Thanks guys, I'm going to be now, its 23:00 and I'm still beat from only getting 4 hours of sleep last night, so I better be off now. If I get it figured out I suppose I owe it to you to post it here. Night.

[Solt]

It's actually very simple. Since you've gone to sleep, here's the answer:

Solve both equations for x^2:

(1) x^2 = 4y^2 - 80y + 340
(2) x^2 = 40 - y^2

Now you can use the associative property, which is the first thing you learn in algebra (If A=B and A=C, then B=C. In this case, A is x^2. This is essentially "substitution.")

(3) 4y^2 - 80y + 340 = 40 - y^2

Order it into standard form:

(4) y^2 - 16y + 60 = 0

Factoring gives you y = 6, y = 10.

Plug the factors back into (1) or (2) to get the corresponding x. It doesn't matter which because both will give you the same answers. Using (2) is easier.

It gives:

y = 6 => x = sqrt(4) = +/-2
y = 10 => x = sqrt(-60) (meaningless, discard this)

Thus, the final answer is (2, 6) and (-2, 6).



Showing it graphically is a bit more complex. Graphing (2) is simple enough, but graphing (1) is a bitch, as you have realized. You have to use a trick called Complete the Square.

First, divide through by 4:

(5) (using (2)) x^2/4 - y^2 + 20y = 85

Look at the quadratic term for y:

(6) y^2 - 20y

Wouldn't it be perfect if we had a "+100" at the end there? Well in fact, we do! Just subtract it from the other side!

(7) x^2/4 - y^2 +20y - 100 = 85 - 100
(8) x^2/4 - (y^2 - 20y + 100) = -15
(9) x^2/4 - (y-10)^2 = -15

(7), (8), and (9) are called Completing the Square. Now, you can easily solve for y and plug it into your graphing calculator to get the graph and see that the intersection of the two equations corresponds to (2,6) and (-2,6).

[hack124x768]

It's actually very simple. Since you've gone to sleep, here's the answer:

Solve both equations for x^2:

(1) x^2 = 4y^2 - 80y + 340
(2) x^2 = 40 - y^2

Now you can use the associative property, which is the first thing you learn in algebra (If A=B and A=C, then B=C. In this case, A is x^2. This is essentially "substitution.")

(3) 4y^2 - 80y + 340 = 40 - y^2

Order it into standard form:

(4) y^2 - 16y + 60 = 0

Factoring gives you y = 6, y = 10.

Plug the factors back into (1) or (2) to get the corresponding x. It doesn't matter which because both will give you the same answers. Using (2) is easier.

It gives:

y = 6 => x = sqrt(4) = +/-2
y = 10 => x = sqrt(-60) (meaningless, discard this)

Thus, the final answer is (2, 6) and (-2, 6).



Showing it graphically is a bit more complex. Graphing (2) is simple enough, but graphing (1) is a bitch, as you have realized. You have to use a trick called Complete the Square.

First, divide through by 4:

(5) (using (2)) x^2/4 - y^2 + 20y = 85

Look at the quadratic term for y:

(6) y^2 - 20y

Wouldn't it be perfect if we had a "+100" at the end there? Well in fact, we do! Just subtract it from the other side!

(7) x^2/4 - y^2 +20y - 100 = 85 - 100
(8) x^2/4 - (y^2 - 20y + 100) = -15
(9) x^2/4 - (y-10)^2 = -15

(7), (8), and (9) are called Completing the Square. Now, you can easily solve for y and plug it into your graphing calculator to get the graph and see that the intersection of the two equations corresponds to (2,6) and (-2,6).

So 4y^2 - 80y + 340 = 40 - y^2
Reduce (divide by 5)
and solve for y, well, I can do that.

Ok. So I was just missing this fist step it looks like. I knew it was something simple.

And I have no idea what I was thinking for the graph. Thanks a lot.

[kira]

Now you can use the associative property, which is the first thing you learn in algebra (If A=B and A=C, then B=C. In this case, A is x^2. This is essentially "substitution.")

This is not the associative property. The associative property has to do with which numbers associate with each other. Such as: (a + b) + c = a + (b + c).

What you are describing is the transitive property.

[Solt]

Now you can use the associative property, which is the first thing you learn in algebra (If A=B and A=C, then B=C. In this case, A is x^2. This is essentially "substitution.")

This is not the associative property. The associative property has to do with which numbers associate with each other. Such as: (a + b) + c = a + (b + c).

What you are describing is the transitive property.

Yea whatever. Same thing. Stupid names, I hate names. I can never remember them. They're not important anyway.

[hermaj]

I would suggest the correct names for things are pretty important, especially when you're trying to teach someone.

Hack! So when are we going to find out how you did?

[hack124x768]

Well, I lucked out on friday and there wasn't enough time for my group to go. So, I guess I'm going on monday. Man I hate talking infront of the class. Oh well. I'll get used to it. I'll need to, for when I rule the world!

[Fernanthonies]

Well, I lucked out on friday and there wasn't enough time for my group to go. So, I guess I'm going on monday. Man I hate talking infront of the class. Oh well. I'll get used to it. I'll need to, for when I rule the world!

try taking a course where you present in front of the class every freaking Tuesday . I sure am glad that's over with.

[Solt]

Well, I lucked out on friday and there wasn't enough time for my group to go. So, I guess I'm going on monday. Man I hate talking infront of the class. Oh well. I'll get used to it. I'll need to, for when I rule the world!

Damn. So I practically did your homework for you.

You better like, do extra problems or something...

[hack124x768]

Well, I lucked out on friday and there wasn't enough time for my group to go. So, I guess I'm going on monday. Man I hate talking infront of the class. Oh well. I'll get used to it. I'll need to, for when I rule the world!

Damn. So I practically did your homework for you.

You better like, do extra problems or something...

Sad thing is, I still don't get it. I'm about ready to start hitting my head against a brick wall. Maybe if I print it and look at it on paper it will make more sense. This sucks. And no, you didn't do the homework for me, As I still have to make a presentation and explain why I did what I did. So I nee to rework it a few times anyways.

Bla.

[Solt]

Well, I lucked out on friday and there wasn't enough time for my group to go. So, I guess I'm going on monday. Man I hate talking infront of the class. Oh well. I'll get used to it. I'll need to, for when I rule the world!

Damn. So I practically did your homework for you.

You better like, do extra problems or something...

Sad thing is, I still don't get it. I'm about ready to start hitting my head against a brick wall. Maybe if I print it and look at it on paper it will make more sense. This sucks. And no, you didn't do the homework for me, As I still have to make a presentation and explain why I did what I did. So I nee to rework it a few times anyways.

Bla.

What part don't you get? Reasons for doing particular steps, or what? We can try to explain more.

[hack124x768]

Well, I lucked out on friday and there wasn't enough time for my group to go. So, I guess I'm going on monday. Man I hate talking infront of the class. Oh well. I'll get used to it. I'll need to, for when I rule the world!

Damn. So I practically did your homework for you.

You better like, do extra problems or something...

Sad thing is, I still don't get it. I'm about ready to start hitting my head against a brick wall. Maybe if I print it and look at it on paper it will make more sense. This sucks. And no, you didn't do the homework for me, As I still have to make a presentation and explain why I did what I did. So I nee to rework it a few times anyways.

Bla.

What part don't you get? Reasons for doing particular steps, or what? We can try to explain more.

Well, printing it and looking at it on paper helped. I think I get it now. The caffeine helped a bit. Thanks a bunch.

[cmacis]

Oh caffeine, the spirit of Pythagoras living on within every mathematician. Because, you know, that's how he got reincarnated.

Though sometimes I would like to turn theorems into tea.

[hack124x768]

66%
That was bad.
I made a few sign errors (on the whiteboard) and we were not in agreement the whole time while presenting. Arguing with teammates (even slightly) docks a lot of points, really fast.

[Gordon]

I'm doing this purely because I'm bored. But I'm also doing this now because my manager can't fathom why I would help some e-stranger with their homework. "No I just don't understand... seriously [laughing] why.. who are these people."


1) x(sqr)-4y(sqr)+80y=340
2) x(sqr)+y(sqr)=40

1) x(sqr)-4y(sqr)+80y-340=0
2) x(sqr)+y(sqr)-40=0

They "collide" at points of intersection.

Set them equal to each other 1=2.
x^2-4y^2+80y-340 = x^2+y^2-40
Group like terms
-5y^2+80y-300
Flip the signs because no one likes negatives
5y^2-80y+300 = 0
Factor
5(y^2-16y+60) = 0
5(y-10)(y-6) = 0
y = 10 or y = 6
Sub into (1), solve for x
x^2-4(10)^2+80(10)-340 = 0
x^2-400+800-340 = 0
x^2+60 = 0
x = sqrt(-60) (not a real answer)
x^2-4(6)^2+80(6)-340 = 0
x^2-144+480-340 = 0
x^2-4 = 0
x = 2 (real answer)

So.. point of intersection is (2,6)


Edit: I assumed you're in highschool, so I solved it at a highschool level.
Edit2: I know you've already been graded on this, but I wanted to contribute.

[Gelsamel]

I had a friend with a similar sentiment. I asked on Physicsforums whether I could explicitly define x(y) if y = x + e^x. And he said to me "So what, some random people are just sitting there waiting to answer your questions? Who are these people!? Why!?".

[Solt]

x^2-4 = 0
x = 2 (real answer)

x also equals -2.

[Gordon]

Flip the signs because no one likes negatives

Pfft.

[The LuigiManiac]

Flip the signs because no one likes negatives

Pfft.

It's true. I like seeing positive numbers more than negative ones.

[hack124x768]

I actually am slightly curious as to what I did wrong. I'll scan my notes/work and post them later.

[Gordon]

The thing you have to remember with most highschool math problems, is they give you all the tools you need to find the solution. Once you get to University (if you so choose) that's when they really start fucking with you.

"Professor I spent all week working on the problem and I couldn't find the solution"

"Very good, there was no solution. Here's this weeks problems."

[hack124x768]

The thing you have to remember with most highschool math problems, is they give you all the tools you need to find the solution. Once you get to University (if you so choose) that's when they really start fucking with you.

"Professor I spent all week working on the problem and I couldn't find the solution"

"Very good, there was no solution. Here's this weeks problems."

Oh god...

[SpitValve]

The good news is that at uni you can use Maple or whatever to skip all the messy little working bits, and just get into the juicy problem itself.