a simple math question hehe

bootymac

Diamond Member
Aug 20, 2001
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0
76
i think i'll post a math questing as well ;)
being the usual retarded, absent minded, retarded, dumbass i am, i actually got it :)

find x and y
x + y = 10
x = y * 0.5

its simply stuff...but o well :cool:

first person to get it wins a "secret" "prize" ;););)
 

bootymac

Diamond Member
Aug 20, 2001
9,597
0
76
dammit!
uh..for your "prize"....move ur mouse in circles 5 times and tell me what happens :)
 

Logix

Diamond Member
Jul 26, 2001
3,627
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d00d, this is basic algebra. Substitute and solve.

y= 10/1.5 = 6.67
x= 5/1.5 = 3.33
 

BruinEd03

Platinum Member
Feb 5, 2001
2,399
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0


<< dammit!
uh..for your "prize"....move ur mouse in circles 5 times and tell me what happens :)
>>



hmm...i moved it in a circle 5 time and a message popped up on my computer saying bootymac would pay me $100 :D

-Ed
 

Noriaki

Lifer
Jun 3, 2000
13,640
1
71
Ima supah bored so I'm going to do it..

x + y = 10
x - .5y = 0

Subtract row 1 from row 2
x + y = 10
0x + -1.5y = -10

divice row two by -1.5

x + y = 10
0x + y = 6.666...

subtract row 2 from row 1

x + 0y = 3.333...
0x + y = 6.666...

Ahh Gauss-Jordan elimination on a 2x2 matrix...gotta love that time wasting ;)
 

AdamDuritz99

Diamond Member
Mar 26, 2000
3,233
0
71
x+y=10
x=y*.5

therefore
x= y/2

y/2 + y = 10

(y/2 + y) = 10

[y(1/2) + y ] =10

y[1/2] = 10

y = 10/(1/2)

y=20


x + (20) = 10
x =-10


it's step by step. I think that's right.


peace
sean

*EDIT*

Well i guess that doesn't work b/c

(-10) + (20) = 10
but

(-10) = (20) * .5
which is

-10 = 10 which is false. :(
 

Noriaki

Lifer
Jun 3, 2000
13,640
1
71
whoa check that answer bad boy.

y = 20, x = -10?

Wouldn't satisfy equation two, -10 = 20/2? Nope



<< [y(1/2) + y ] =10

y[1/2] = 10
>>



That should be y[3/2] = 10, not y[1/2] = 10.

;)
 

AdamDuritz99

Diamond Member
Mar 26, 2000
3,233
0
71


<< whoa check that answer bad boy.

y = 20, x = -10?

Wouldn't satisfy equation two, -10 = 20/2? Nope



<< [y(1/2) + y ] =10

y[1/2] = 10
>>



That should be y[3/2] = 10, not y[1/2] = 10.

;)
>>




aww, you wise one, that's where i missed it. oh well. If i did this on paper i would of got it right...i think ;)


peace
sean
 

y is (20/3) and x is (10/3)!!

Work:

Eq 1: x+y = 10
Eq 2: x = 0.5y

We can immediately see that x = (1/2)y.

Using the substituion method, we replace x with (1/2)y in Eq 1.

Thus, (1/2)y + y = 10

==> (3/2)y = 10
==> y = 20/3

We now know the value of y, so we can find x: x = (1/2)y = (1/2)*(20/3)= 10/3!

You could also use the Gaussian elimination method. But the substitution method is the fastest here . . . for me, at least!

Simple, eh! Hehehe! ;)

Ugh! My usual typos!
 

Noriaki

Lifer
Jun 3, 2000
13,640
1
71


<< You could also use the Gaussian elimination method. But the substitution method is the fastest here . . . for me, at least! >>

lol of course. For a 2 equation 2 variable, Gaussian Elimination is massive overkill, substitution is by far the smarter way to do.

But Ima Supah Dupah bored!



<< Simple, eh! Hehehe! >>

Eh? You some kinda canadian?

You gotta 'puter in ur igloo?

(Kidding, I'm Canadian to ;))
 

Mday

Lifer
Oct 14, 1999
18,647
1
81
when given in that form, direct substitution is obvious.

since x = y/2

x + y = y/2 + y = 10 = 3y/2
y = 20/3
x = 10/3