Solve for X

[Chruser]

I put together a couple of equations for you not to actually solve for, but to break out X from. Some are first-degree equations, some are second-degree, and the difficulty varies quite a bit. Remember to post the solutions step by step as it's extremely easy to just cram the equations into an equation solver/breakout otherwise:


[list=1]
5x(75x+32y)=5xy

(x+y)/(y+x)=xy

(x-y)/(y-x)=xy

sqrt(5x)=y

(x+y)/(x-y)=2xy

(1337x)^2-(15y)^2=sqrt(y)
[/list=1]

[Titusfied]

Well, I'm not doing these now, no time, but:

4. x = (y^2)/5

[Chruser]

Yeah, that answer is correct, but do post the steps you managed to solve it through.

[Demosthenes]

#4

sqrt(5x)=y

u square both sides to get rid of the sqrt()

u get 5x=y^2

u divide 5 on both sides to isolate x

x=(y^2)/5



#1 (i have a feeling im wrong so plz tell me if i am...if im not ill post step by step)

x=sqrt((-27xy)/375)

[Chruser]

4 is right.

I don't know what on earth you did with #1, but here's some basic hints for it:

5x(75x+32y)=5xy
375x^2+160xy-5xy=0
x(375x+155y)=0

So it's a second-degree equation, and one x has to equal 0, but see if you can figure out what the other one equals.

[Demosthenes]

lol...im an idiot...sorry

o well...i tried...although ill work through all of them when i have a chance eventually

[Xenn]

1. 15+45=60
2. 60=60

State the reason that makes going from step 1 to step 2 true.

Easy shiz

[RoboticSilence]

Number 6!
This one is a bitch...
Had to resolve to paintchat do it right.
http://www.zelaron.com/gear/equationhell.jpg

[Chruser]

Oops, 255 should be 225 isntead, except for that it's all good.

[quikspy67]

It's the weekend I don't wanna do this.

[RoboticSilence]

Too bad... you need the practice... it's to excite the brain.

[J_iceman]

this is algebra right???

4. sqrt(5x)=y

1. 5x=y/sqrt
2. x= 5y/sqrt

so is the answer THAT or x=y/5xsqrt

??

[Demosthenes]

sqrt is an operation not a variable

sqrt stands for square toor

[Chruser]

Ugh, #5 is giving me a headache, I can't crack it.
y=(x-y)(2xy)-x is about as far as I get, and it definitely seems like it's gonna be a fourth-grader, so feel free to help me out with this one a bit.

[Chruser]

Bump. Anyone care to give #5 a try?

[tidus2005]

ill try it but i dont know what ill come up with.

[LordZpider]

5x(75x+32y)=5xy
375x^2 + 160xy = 5xy
375x^2 = -155xy
375x = -155y
x = -155/375y

(x+y)/(y+x)=xy
(x+y)/(x+y)=xy
1=xy
x=1/y

(x-y)/(y-x)=xy
(x-y)/ -(x-y)=xy
-1=xy
x=-1/y

sqrt(5x)=y
sqrt(5x)^2=y^2
5x=y
x=y/5

(x+y)/(x-y)=2xy
-(y-x)/(x-y)=2xy
-[(y-x)/(x-y)]=2xy
-(-1) = 2xy
x= 1/2y

(1337x)^2-(15y)^2=sqrt(y)
I think your solution to this is correct, so no point in posting

[Chruser]

LordZpider:

#1 should have the y in the numerator rather than the denominator. Oh, and the other root is 0.

#2 is correct.

#3 is correct.

#4 is correct.

#5 is incorrect. You can't turn (x+y)/(x-y)=2xy into
-(y-x)/(x-y)=2xy, trying to break out -1 would result in -(-y-x)/(x-y)=2xy.

[RaZ]

hmm for number 5 i got

X = 2x^2 y - 2xy^2 - y

[LordZpider]

yea, for #1 it's hard to write it with one line but it's (-155/375)y

and yes, my #5 is wrong, i have a full load of classes today, and as a matter of fact i'm going to "matrix theory" right now, i did that in sorta a rush (my excuse, obviously), so i'll try to work on it later.

It's sorta embrassing as an engineer to get algebra problems wrong, then again nasa crashed into mars cuz they forgot to convert :-P.

Cheers.

[LordZpider]

back from class, here's solution...

#5
(x+y)/(x-y)=2xy
long division
1 + (2y/(x-y))=2xy
multiply everything by (x-y)

HANG ON

then you seperate stuff and quadratic, i'll finish it later

[Titusfied]

Well shit, now I'm fucking pissed.... Well, I was in my night class, bored as usual, so I decided to tackle that 5th equation. Well, as it turns out, I did the wrong equation, but the process still works for this one, which I will not do at this time again.... Alright, I'll admit, I did kinda cheat.... Chruser said that is had to be turned into a fourth order polynomial, so with that in mind:

[Chruser]

That's really close Titus, it's not entirely correct, but that's definitely on the right track. That solution was scarily long too.

[Titusfied]

Bah, I guess I'll work on the right equation today. And yes, it is scarily long...

[Chruser]

This is what my symbolic solver gives me:

http://www.zelaron.com/gear/solu.jpg

[LordZpider]

http://www.ews.uiuc.edu/~cngan/math.jpg

inside the square root, are you sure it's +12y^2, chruser? cuz I got -12y^2. Everything else is the same, in different order.

[OmniFalcon]

for #1 i got

5. x = square root of 155xyi over 5 square root 15

:-/ il stick to algebra 2 >.<

[tidus2005]

dam LordZpider i wish i was as smart as you lol i look at your work and cant even follow it lol. and that freaks me out becuase i am in my schools highest math class lol are you in high school or college?

[OmniFalcon]

i wish i had brains like dat =-0

[Titusfied]

Well, obviously he cut out a shit load of steps. LordZpider, please write out your steps so we can follow it next time. I am fine up until you wrote 0 = "B L A H".... Then you basically solved it after that. Write your steps after that.

You definitely skipped a lot of steps, and there is no way anyone can just intuitively solve the way you did from that one step to the next...

[RoboticSilence]

Why did you even bother writing x^0?

[LordZpider]

Tidus, I'm in college as you can obviously tell in my sig.

Robot, the reason for writing x^0, cuz i'm afriad people might not follow that i put it in quadratic form.

And I'm going to write how to follow this...

Line 1->2: Long Division
Line 2->3: multiplied every element with (x-y)
Line 3->4: on the left side added; on the right side distribed
Line 4->5: put x on right side and use communitive property to group the first 2 terms
Line 5->6: subtracted y from both sides; x^0 = 1 so i multiply by 1, which doesn't change the answer; also reverse distrubted (is there a better way to say this?) x into 2y^2-x.
Line 6->7: Quadratic formula (Algebra I)
Line 7->8: distributed and simplified inside the root and the negative outside the root
Line 8->9: added

Oh if you guys need help with high school math up to about 2nd year college math(you only need math as an engineer or math major), i can TRY to help you. Being a nice guy and all... :)

[Titusfied]

Heh, if you did long division for step 1 -> 2, that was quite unnecessary. Just break up the numerator into x - y + 2y, then get to step 2.

Ahhh, didn't even notice Quadratic Formula.... I should have too.

Oh, and how did you justify my questions with "I'm in college, as you obviously tell from my sig,"? That makes no sense what-so-ever. I too am in my 4th year of college, and only 1 year away from getting my BS and MS in Power Engineering. Math is my strong point.

[LordZpider]

What i meant was that my signature has my college on it, what's power engineer? where are you at?

And speaking of things that don't make sense, "1 year away from getting BS and MS," you can't one year from BOTH, i mean you need your undergraduates degree to even go into grad school and get a MS.

Oh this isn't suppose to be offensive, tho proof-reading it, i realized it sounded sorta anal.

[Titusfied]

Power Engineering is a track of Electrical Engineering. I go to Drexel University, and they have a program that you must get accepted into and maintain 3.25 GPA to stay in that enables the student to graduate with a BS and MS in five years.

[LordZpider]

wow that's a really good program, i'm jealous