# hmmm... just wondering

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• Sep 4th 2006, 06:33 PM
dan
hmmm... just wondering
is it posible to solve a non-liner system...and how!!!
for example
$\displaystyle x^2 +y^3 =10$
$\displaystyle x^3 -y^4 =-6$

if any one has any thoughts on how you do this please let me know

~dan
• Sep 4th 2006, 06:36 PM
ThePerfectHacker
Quote:

Originally Posted by dan
is it posible to solve a non-liner system...and how!!!
for example
$\displaystyle x^2 +y^3 =10$
$\displaystyle x^3 -y^4 =-6$

if any one has any thoughts on how you do this please let me know

~dan

Of course,
$\displaystyle x^3=y^4-6$
$\displaystyle x=\sqrt[3]{y^4-6}$
Substitute,
$\displaystyle (y^4-6)^{2/3}+y^3=10$
Thus,
$\displaystyle (y^4-6)^{2/3}=(10+y^3)$
Cube both sides,
$\displaystyle (y^2-6)^2=(10+y^3)^3$
Expand and solve for the nasty nontic.

Note, the solutions not necessarily can be expressed in algebraic terms.
• Sep 5th 2006, 06:34 AM
dan
Quote:

Originally Posted by ThePerfectHacker

$\displaystyle (y^2-6)^2=(10+y^3)^3$
Expand and solve for the nasty nontic.

i hate to ask...but how do you slolve for y in that equation.
it seems i'v spent hours working on probelms of that type before....
would i use the "binomal therem"?
dan
• Sep 5th 2006, 07:17 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Of course,
$\displaystyle x^3=y^4-6$
$\displaystyle x=\sqrt[3]{y^4-6}$
Substitute,
$\displaystyle (y^4-6)^{2/3}+y^3=10$
Thus,
$\displaystyle (y^4-6)^{2/3}=(10+y^3)$
Cube both sides,
$\displaystyle (y^2-6)^2=(10+y^3)^3$
Expand and solve for the nasty nontic.

Note, the solutions not necessarily can be expressed in algebraic terms.

using:

$\displaystyle x=\pm \sqrt{10-y^3}$

and substituting into

$\displaystyle x^3 -y^4 =-6$

will give two octics - a much easier problem than a single nonic. :D?

RonL
• Sep 5th 2006, 07:20 AM
CaptainBlack
Quote:

Originally Posted by dan
i hate to ask...but how do you slolve for y in that equation.
it seems i'v spent hours working on probelms of that type before....
would i use the "binomal therem"?
dan

If you are lucky it will have a special form that will allow you to find
all or some or the roots. Otherwise numerically.

RonL
• Sep 5th 2006, 07:24 AM
CaptainBlack
Quote:

Originally Posted by dan
is it posible to solve a non-liner system...and how!!!
for example
$\displaystyle x^2 +y^3 =10$
$\displaystyle x^3 -y^4 =-6$

if any one has any thoughts on how you do this please let me know

~dan

If you are happy with approximate solutions one could start garphically.

These two equations represent curves in the x-y plane. Sketch them
and your first approximation to the soution/s will be the point/s of
intersection.

Then they can be refined with an appropriate numerical procedure (say
linearising about the approximate roots, or something like that).

RonL
• Sep 5th 2006, 07:46 AM
ThePerfectHacker
Quote:

Originally Posted by dan
i hate to ask...but how do you slolve for y in that equation.
it seems i'v spent hours working on probelms of that type before....
would i use the "binomal therem"?
dan

Not necessarily there is a general method.
Yes, you do expand with binomial theorem then work with polynomial equation.
(Do you not just love it when mathematicians say "I cannot solve the problem but I can show it cannot be solved!")
• Sep 5th 2006, 09:05 AM
dan
Quote:

Originally Posted by ThePerfectHacker
(Do you not just love it when mathematicians say "I cannot solve the problem but I can show it cannot be solved!")

hahaha
• Sep 5th 2006, 09:12 AM
dan
so if i had a polynomal. could i solve with newton??

i dont so how you get a nontic though $\displaystyle (10 +y^3)^3$ just because 3*3 is nine???

some how i got from $\displaystyle (y^2 -6)^2 =(10 +y^3)^3$

to $\displaystyle 0= y^5 + 59y^4 +90y^3 + 12y^2 + 964$

is that right???
dan
• Sep 5th 2006, 09:15 AM
dan
Quote:

Originally Posted by CaptainBlack
using:

$\displaystyle x=\pm \sqrt{10-y^3}$

and substituting into

$\displaystyle x^3 -y^4 =-6$

will give two octics - a much easier problem than an nonic. :D

RonL

mabye i'll try that
• Sep 5th 2006, 09:37 AM
ThePerfectHacker
Quote:

Originally Posted by dan

some how i got from $\displaystyle (y^2 -6)^2 =(10 +y^3)^3$

$\displaystyle y^4-12y^2+36=1000+300y^3+30y^6+y^9$
Thus,
$\displaystyle y^9+30y^6-y^4+300y^3+12y^2+9964=0$
• Sep 5th 2006, 09:48 AM
dan
Quote:

Originally Posted by ThePerfectHacker
$\displaystyle y^4-12y^2+36=1000+300y^3+30y^6+y^9$
Thus,
$\displaystyle y^9+30y^6-y^4+300y^3+12y^2+9964=0$

yea...thanks i did something wrong...what else is new...
• Sep 5th 2006, 09:58 AM
dan
so if
$\displaystyle f(y)=y^9 + 30y^6 - y^4 +300y^3 + 12y^2 +9964$
then
$\displaystyle f'(y)=y(9y^7 +180y^4 -4y^2 +900y +24)$

so could i use newton for here to approximate [tex] y [tex]
• Sep 5th 2006, 10:03 AM
dan
by my calc $\displaystyle y \approx -3.13425651413923$

look right??
• Sep 5th 2006, 01:28 PM
CaptainBlack
Quote:

Originally Posted by dan
by my calc $\displaystyle y \approx -3.13425651413923$

look right??

What do you get if you substitute this back into the original equations?
Is there a x which with this as a value of y solves the equations?

RonL

(when I do this I get no consistent solution for x, but I could be doing the
arithmetic wrongly :( )
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