Thread: hmmm... just wondering

1. Originally Posted by CaptainBlack
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 )
Using the graphical method I find an initial estimate for the solution
is x=2.36, y=1.64, and there is probably but one real solution.

RonL

2. hmmmm.....

3. Originally Posted by ThePerfectHacker
$(y^2 - 6)^2 = (10+y^3)^3$
Thus,
$y^9+30y^6-y^4+300y^3+12y^2+9964=0$
is this correct...for some reason it's not getting me a good value for y...

i'm getting tired of this problem...
dan

4. Originally Posted by CaptainBlack
Using the graphical method I find an initial estimate for the solution
is x=2.36, y=1.64, and there is probably but one real solution.

RonL
what is the graphical method ??

dan

5. Originally Posted by dan
what is the graphical method ??

dan
Plot a graph of the curve represented by:

$
x^2 +y^3 =10
$
,

and on the same axes plot the graph the curve represented by:

$
x^3 -y^4 =-6
$
.

The points where these curves cross are roots of the system.

RonL

6. Originally Posted by dan
what is the graphical method ??

dan
You draw the graw and see where it interesects the x-axis.
There is only one such place.
Thus, there is only no real root.

7. Originally Posted by ThePerfectHacker
You draw the graw and see where it interesects the x-axis.
There is only one such place.
Thus, there is only no real root.
Try the graphs of the original two curves, which do seem to have a root.

RonL

8. Hier

9. Originally Posted by ThePerfectHacker
Hier
That seems more accurate than my initial sketch, but now we have an
estimate we can replot about the solution.

RonL

10. Originally Posted by ThePerfectHacker
$y^4-12y^2+36=1000+300y^3+30y^6+y^9$
Thus,
$y^9+30y^6-y^4+300y^3+12y^2+9964=0$
Typo, last should be:

$y^9+30y^6-y^4+300y^3+12y^2+964=0$

RonL

11. Originally Posted by ThePerfectHacker
Of course,
$x^3=y^4-6$
$x=\sqrt[3]{y^4-6}$
Substitute,
$(y^4-6)^{2/3}+y^3=10$
Thus,
$(y^4-6)^{2/3}=(10+y^3)$
Typo, last should be:

$(y^4-6)^{2/3}=(10-y^3)$

This canges everything down stream along this line of solution

RonL

12. Originally Posted by CaptainBlack
Typo, last should be:

$(y^4-6)^{2/3}=(10-y^3)$

This canges everything down stream along this line of solution

RonL
Expanding this gives:

$
y^9 + y^8 - 30 y^6 - 12 y^4 + 300 y^3 - 964=0
$

which may be solved numerically to give $y = 1.873782741$

RonL

13. Originally Posted by CaptainBlack
Expanding this gives:

$
y^9 + y^8 - 30 y^6 - 12 y^4 + 300 y^3 - 964=0
$

which may be solved numerically to give $y = 1.873782741$

RonL
captblack, you're the man you get a +rep.
thanks guys...case closed...untill later

dan

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