Algebra minigame

[xinjinbei]

ask random algebra questions or answer them (by quoting post)

I start


2+2

[Homer43]

lack of directions

[Sam]

Homer43 is right. Some algebraic person you are.

You fail.

[Diamond187]

That's not even totally algebraic. It lacks an equal sign and a variable. That's just an incomplete statement.

[Liokae]

It lacks an equals sign, a variable, *and* directions, such as "Solve for x."

[Sam]

Thank you Liokae, for restating the previous 3 posts.

It lacks an equal sign and a variable.

lack of directions

:-D

[Doug05257]

It is not an algebraic equation, anyway. It is an arithmetic equation, as it has no variable, equal sign or inequality, and as such cannot be answered algebraically. 2+2=4, by the way. Here's a better algebraic equation.

Solve for each of the variables.

A=Y
Y=X
Y=Z
Z=1


23X+(-46Z)+92Y=69A

[Diamond187]

Um, they're all 1...

You don't even need the one at the bottom...

[boomboom14]

yea really... how about this.

solve for w, x, y, and z

w*x=11y+2z
4y+2z=2w
3x=2w+z
2x+w+y=20z

good luck!
and if u think u have it pm me the answer and ill announce the first winner, dont just post it.

[Diamond187]

Well, that's a pain. Way to make it non-linear. There are probly multiple solutions. *sigh, goes to work anyway*

[boomboom14]

its not that hard tho...just remember the substitution property of equality...lol...

[JoeyG]

What grade is that? I'm in Grade 10 algebra, and that looks pretty easy, but is there some trick to solving it quicker?

[Diamond187]

Matrices! At least, for the linear equations. But even that isn't that much faster than just manipulating the equations. I keep getting really ugly answers so I just gave up.

[boomboom14]

ok ill give u a hint...all of the answers are integers...just use the substitution property of equality...if you know what that means...im in geometry and i carefully made this so it works out and matrices dont work...if i had made it linnear it wouldve been too easy...and this is algebra 1...well they didnt teach 4 variable equations but with the principles you learn there you can do it...

[Diamond187]

Matrices should still work for all linear equations, ie. the last 3. From there you get everything in terms of one variable and solve the non-linear equation. Problem is, the linear part turns ugly. I coulda just screwed up some arithmatic, I'm not trying that hard.

[boomboom14]

ITS NOT LINEAR!!!!!

[Diamond187]

YOU STILL HAVE 3 LINEAR EQUATIONS!!!!

I'm not a complete idiot. While you can't solve for all 4 using a matrix you can get every other variable in terms of one and then get a single variable, non-linear expression.

[boomboom14]

w*x=11y+2z
4y+2z=2w
3x=2w+z
2x+w+y=20z

Ok, you tell me which ones of those are linear. I can't remember the specific equations because I learned it a year ago but still. None of them look linnear to me.

[Leperkawn]

Thats not a very nice problem.

*goes to figure out*

[DarkReality]

I just spent a whole lot of time solving those. Now take a second look at those equations.

w = 0, x = 0, y = 0, z = 0.

:-D

I ended up with something like -7410z = 0 and suddenly realized there are no constants. Since there are 4 equations and 4 variables, I don't think there are multiple answers, despite the first equation being a bit odd.

A linear equation is one which follows the basic outline of:

ax +by + cz = d

where a, b, c, and d are constants and x, y, and z are variables. However, I have no clue if multiplying two variables with one another still counts as linear, as that's a type of equation I've never had to deal with when doing linear equations.

[boomboom14]

yea yea. thanks for reminding me of the linear equation but it is Ab+By=C and one of the other forms is Y=mX+b. but yea the 0,0,0,0 thing is right and i was wondering who would figure out the other answer.

[DarkReality]

So there is another one? Oh joy.

A linear equation can have as many variable as you want. Technically au+bv+cw+dx+ey+fz = g is a linear equation. An incredibly ugly one that you'll most likely never need or even see, but it still counts as a linear equation, as it has no exponents.

Oh my god that is nasty to solve using substitution.

[Diamond187]

If you multiply two variables in an equation it isn't linear, since you can write both variables in terms of one other you'll invariably end up with a squared term. That's probably where the other answer comes from.

[boomboom14]

oh shoot...

i just realized that when i made those i forgot to put any numeric values in so just take one of the equations out and put this one in... it doesnt matter which one...

put this 1 in...

3w+5x=44+y

[DarkReality]

Didn't I just say that? Did you even read my post?

but yea the 0,0,0,0 thing is right and i was wondering who would figure out the other answer.

And didn't you intentionally do that, according to your post further up?

You scare me, good sir. A lot.

[thorcat1391]

solve this one for x

13x² + 54x - 234 = 0

this one is harder

[DarkReality]

Not really...

x1 = -6.801...
x2 = 2.6468...

[thorcat1391]

there is only one x. the one with a two is x squared.

[DarkReality]

... a quadratic equation can have mutlple answers. In fact, it can have as many answers as the highest exponent, in this case 2, as you had x².

What I meant with x1 and x2 is that there are two possible answers for x. Normally you'd write them as an index (small number on the lower right corner of the x) but that's a bit difficult without html.

[Diamond187]

Alright, try this:

From the base pair of vectors (1,3,0) and (-2,0,1) give me three orthogonal (perpendicular), unit vectors which can be used to define a 3 dimensional coordinate system.

Not for beginners but not that hard if you know what I'm asking. (Hint: there are two answers)

[DarkReality]

So let me get this straight, you want three orthogonal vectors that are made up of multiples of your base vectors? Sorry, I'm used to all this in german >_< It's a bit late now, but I'm sure I'll have it my tommorow if no one else does.

[thorcat1391]

oh im sorry, i didnt realize that you meant that they were the plus and minus awnsers. i am also sorry that i was not on to avoid all this confusion.

[Diamond187]

So let me get this straight, you want three orthogonal vectors that are made up of multiples of your base vectors? Sorry, I'm used to all this in german >_< It's a bit late now, but I'm sure I'll have it my tommorow if no one else does.

They won't all be multiples of the base vectors, that would be contradicting orthogonality. Plus, I want unit vectors (ie. length of 1). I'll give you this hint though: The two vectors aren't orthogonal to each other, so only one of them will appear in your final answer (and that's why there are two answers).

[boomboom14]

is this algebra 1 stuff? bc im not in algebra 2...

[Diamond187]

Heh, it's linear algebra stuff. I'm fairly confident that isn't covered in algebra 1 or 2 (I'm assuming those are American highschool terms).

[boomboom14]

yea, sry i forgot that other countries call it different stuff...well ok i haven't learned 3d figures b4...