#1
Yay, i come here to ask the pit for help. I know there are some smarty pants lurking in the shadows.

so here it is! (^ means squared here)

x^ + y^ = 16
y = 3x - 1

I got as far as
x^+ (3x-1)^ = 16
x^ + 9x^ - 6x +1 = 16
10x^-6x - 15 = 0
(something x^ -/+ ) (something x +/- )

^^^^^ Factorise? I can't work out what numbers to put in. So i tried the forumla

-b +/- √b^-4ac
----------------------
2a

and suprise suprise got nowhere apart from....
+/- 4.81 (2.d.p)
+/- 7.19 (2.d.p)

???????? Yeah, i have a whole 4 pages of **** like this (all involving simultaneous equations) so get ready for cooly fun time.
Any help is greatly appreciated thanks.
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Quote by Miggy01
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#3
lol, i can do t3h easy ones. But what the hell is that. Its only the numbers in the factorising that i can't work out. And i don't know if the forumla answers are right, or where to put the two numbers it gives. :S
Member of the 'Dr.Cox is my Mentor' group

Quote by Miggy01
I was kicking a balloon around, and kicked the back of my other foot.
I broke my toe as a result.
#5
The answer is Pi

It's always Pi
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#6
Quote by meh!
The answer is Pi

It's always Pi

haha, I wish.

I'm doing Math's at Uni Level now and it's ridiculously hard.

I'll put some of my questions up soon

EDIT:

{xEZ | x2<16} = {-3,-2,-1,0,1,2,3} = {xEZ | |x|<4}}

Another EDIT:

{x,yEZ|-2<x<3^0<y<xo(x,y)} = ?
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Last edited by Kingyem0c0re at Oct 27, 2007,
#7
i'm pretty sure your numbers in the (5x-na)(3x+nb) factorization are going to be decimals.
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#8
The last equation doesn't factor, and I got 1.561 and -0.961 for X. (My calculator doesn't lie )

So that makes Y=3.863, -3.883

Its just one math problem, if its a real pain, move on.
#9
x^2 + (3x-1)^2 = 16
x^2 + 9x^2 - 6x + 1 = 16
10x^2 - 6x - 15 = 0

now plug these coefficients into the quadratic equation and you should get the answers from the last post: a = 10, b = -6, c = -15.

x = [-(-6) + sqrt(6^2 - 4*10*(-15))]/(2*10) = 1.56095

x = [-(-6) - sqrt(6^2 - 4*10*(-15))]/(2*10) = -0.960952

I'm sure you can get the y-values from here!
Last edited by bentfocus at Oct 27, 2007,
#10
why cant you just take the square root of x^2 + (3x-1)^2 = 16?

x + (3x - 1) = 4
4x= 5
x = 5/4

of course, that may not be how youre supposed to do it. youre using a technique i did 2 years ago and i dont use in calculus..
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#11
Quote by KickOutTheJamsX
why cant you just take the square root of x^2 + (3x-1)^2 = 16?

x + (3x - 1) = 4
4x= 5
x = 5/4
.


Well, the square root of x^2 + (3x-1)^2 = 3.16228*sqrt(x^2 - 0.6x + 0.1),
not: x + (3x - 1).
If you did that it's like saying sqrt(4 + 4) = 2 + 2 = 4, when it is clearly
sqrt(4 + 4) = sqrt(8) = sqrt(2*4) = 2*sqrt(2)
#12
Wait lol i just realised i put the numbers into the forumla wrong.
(-b) would be (+6) because the b is -6 already.

I eventually, have done the whole paper. Thanks guys!
Member of the 'Dr.Cox is my Mentor' group

Quote by Miggy01
I was kicking a balloon around, and kicked the back of my other foot.
I broke my toe as a result.
#15
Completing the square is so much easier.

10x^2 - 6x - 15 = 0

x^2 - 6/10x - 15/10 = 0

(x - 3/10)^2 - 9/100 - 15/10 = 0

(x - 3/10)^2 = 159/100

x - 3/10 = +/-root(159/100)

x = 3/10+/-root(159/100)
#16


Hope that helps.
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#17
I might sound like an idiot but can you not just do a square root job on the top equation so you get x+y=4?.....No..?

EDIT: Yeah, the guy above me is right, substitution should work
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Last edited by con job at Oct 28, 2007,
#18
^^^we have a winner

except b = -6 not 6

btw thats not hard

try uni maths mate you'll cry
Last edited by henza_x at Oct 28, 2007,
#19
Quote by con job
I might sound like an idiot but can you not just do a square root job on the top equation so you get x+y=4?.....No..?

EDIT: Yeah, the guy above me is right, substitution should work



3^2 + 4^2 = 5^2

3 + 4 =/= 5
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#21
Quote by JamieB
3^2 + 4^2 = 5^2

3 + 4 =/= 5


Probably should have checked did my theory make sense before I said it
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#22
Quote by _Tenacious_
Yay, i come here to ask the pit for help. I know there are some smarty pants lurking in the shadows.

so here it is! (^ means squared here)

x^ + y^ = 16
y = 3x - 1


How old are you? Cause we saw this when I was 14 or 15, it's easy.
If you're not sure about it, check with your calculator (Texas if you got one)
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#23
Quote by con job
I might sound like an idiot but can you not just do a square root job on the top equation so you get x+y=4?.....No..?


dont think so... when you route them all, the 2 can be positive or negative