Help me win cute girl @ work brownie points. (Math question)

[fitz19d]

Girl struggling with some algebra. Been too long for me. Trying to find x y and z from the following.

3x-y+2 = 11
2x-y-42 = 22
2x-4y+z = 13


I had through loose rough figuring that is probably wrong worked out to x = 55 and stuck on 97 + z = 4y or Alternatively 110-4y= 13-z. (EDIT) Alright now think I'm completely off myself. Per one of my college prof. "We're criminal justice, we don't do math."

She had some online thing that allegedly made it x = 4 y = -2 and z= -3. (Feel like that's wrong but again it's been 10 years and I hated math)

[Flatline]

x=-55
y=-174
z=-573

To solve subtract the second equation from the first and solve for x. They use x to solve for y in the first equation. Finally use x and y to solve for z in the third equation.

And be careful with those online things, they usually aren't based on algebraic solving but instead on approximations so they can be wrong in some cases.

[fitz19d]

Funny, after wasting a while, and getting about half of what you had, figured out handwriting struck.

Thought her Z were 2's.

So was actually
3x-y+z=11
2x-y-4z=22
2x-4y+z=13

Fun bringing back the old memories and frustrations. Sad part was generally people went to me for help at the time. I always used abstract math and ended up with the answer but couldn't show my work to save my life. Idiot savant?



Edit: And you gave me the right method to use so I find the right answers for the correct equations. Thanks very much.

x=-55
y=-174
z=-573

To solve subtract the second equation from the first and solve for x. They use x to solve for y in the first equation. Finally use x and y to solve for z in the third equation.

And be careful with those online things, they usually aren't based on algebraic solving but instead on approximations so they can be wrong in some cases.

[bryjcom]

This is a very long math problem. It involves back and forth and back and forth of the equations. This might help you
http://www.algebra.com/algebra/homework/coordinate/Linear-systems.faq.question.148339.html

[bryjcom]

Started doing work on the equations. So far I got y= -2

I'll be back in a bit as soon as I solve for the others.

[bryjcom]

x=4

[fitz19d]

Save the effort if you see this, edited above to say I'd gotten it.

Kinda puts me in the mood for some more but she's off shift now and the math problems with her.

[bryjcom]

z=-3

[bryjcom]

Save the effort if you see this, edited above to say I'd gotten it.

Kinda puts me in the mood for some more but she's off shift now and the math problems with her.


Its all right, I had fun doing it. Seems like I got the same numbers she got.

Good luck with the girl.

[bigshane]

when I'm checking the kid's homework, and I find myself in the more and more familiar land of the brain-fart, www.wolframalpha.com/ (http://www.wolframalpha.com/) saves me.

[Sharpienads]

Save the effort if you see this, edited above to say I'd gotten it.

Kinda puts me in the mood for some more but she's off shift now and the math problems with her.

Doing math puts you in the mood? Weirdo.

I've been doing my wife's math homework and even though it is a pita, it is kinda fun.

[Danimal]

If you have a graphing calculator it will take you about 2 seconds to solve using a matrix. Just a sec and I post how.

[Danimal]

Here is a link to the method, just do a 3 x 3 and 3 x 1 matrix. This is the only way to go for dynamics where you have 5 variables or more and 5 or more equations. That would take hours to solve.

http://www.mathplanet.com/education/algebra-2/matrices/using-matrices-when-solving-system-of-equations

But for your case your matrix looks like:

[3 -1 1 ] [x] = [11]
[2 -1 -4] [y] = [22]
[2 -4 1 ] [z] = [13]

Now the 3 x 3 is matrix a or [A] and the last 3 x 1 is matrix b or [B]. Plug those into your calculator then push [A]^(-1)*[B].

The answer you get is:

[4 ] x = 4
[-2] y = -2
[3 ] z = 3

Just in case she asks you any more questions in the near future you will look like a genius when you have the answer in seconds.

[BPTactical]

Don't get your meat where you get your bread............

[kwando]

I feel stupid

[wreave]

I hope the brownie points work out for you, but keep this in mind for next time:

http://www.forbes.com/sites/sap/2011/11/16/do-me-a-favor-so-youll-like-me-the-reverse-psychology-of-likeability/

[Monky]

This was far too early on a saturday morning.. my brain hurts.

[BuffCyclist]

Here is a link to the method, just do a 3 x 3 and 3 x 1 matrix. This is the only way to go for dynamics where you have 5 variables or more and 5 or more equations. That would take hours to solve.

http://www.mathplanet.com/education/algebra-2/matrices/using-matrices-when-solving-system-of-equations

But for your case your matrix looks like:

[3 -1 1 ] [x] = [11]
[2 -1 -4] [y] = [22]
[2 -4 1 ] [z] = [13]

Now the 3 x 3 is matrix a or [A] and the last 3 x 1 is matrix b or [B]. Plug those into your calculator then push [A]^(-1)*[B].

The answer you get is:

[4 ] x = 4
[-2] y = -2
[3 ] z = 3

Just in case she asks you any more questions in the near future you will look like a genius when you have the answer in seconds.

This, I was going to pull out Matlab and solve it this way, but it looks like someone beat me to it, and I'm tired.

In Matlab though, this is how simple it would be.

A = [3 -1 1 ]
[2 -1 -4 ]
[2 -4 1 ]

b = [11]
[22]
[13]


Soln = A\b

Thats a left divide, not a regular divide. Very easy using this method and I'm sure a graphing calculator is just as easy (if you know how to use it lol).

[MrPrena]

It is better to just use a damn linear algebra for those equations.

I think Danimal already covered it above.

[bryjcom]

You pussies! I used 3 pieces of paper and my god given brain to solve those equations... Thats what you get by trying to make me look stupid with an easier way to solve a math problem!![Tooth]

[hghclsswhitetrsh]

I love lamp.

[bogie]

More importantly, do the brownie points now [legally] include pot?