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Solving 2 equation with 2 variables (hard)

 
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Apr1-10, 06:00 AM   #1
 

Solving 2 equation with 2 variables (hard)




I really need some help here (no program is allowed )
 
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Apr1-10, 06:02 AM   #2

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What work have you done on the problem so far?
 
Apr1-10, 07:15 AM   #3
 
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I'm not sure why you would consider that "hard". You can solve the second equation for x as a quadratic function of y. Putting that into the first equation gives a cubic equation for y.
 
Apr1-10, 07:40 AM   #4
 

Solving 2 equation with 2 variables (hard)


The problem is to find exact values, not estimatet values. Therefore: do not use any programs...
 
Apr1-10, 07:46 AM   #5
 
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Indeed. Do you have any strategies to find a root? There are many you can try. Once you find one, you can divide and just solve a quadratic.
 
Apr1-10, 07:57 AM   #6
 
Here is what I have done so far:








But these solutions contains "cos". It is possible to rewrite the equations to the formula

and then find solutions without any trigonometrical components.

That is my problem...
 
Apr1-10, 08:09 AM   #7
 
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Would have been a simpler computation for the program to have changed the x into 1/y rather than the other way around, small numbers to deal with overall. Were you expected to solve this cubic analytically and it's roots actually looked like that?
 
Apr1-10, 08:18 AM   #8
 
Yes. It does not matter how you start. You will get the same solutions with the same trigonometrical components. Plz help me find solutions without any trigonometrical components.
 
Apr1-10, 10:16 AM   #9
 
There is a typo in the problem: "79" should be "78".
 
Apr1-10, 10:48 AM   #10
 
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Is it just because the answer drops our nicely if that is the case, or do you somehow know it as a fact :S ?
 
Apr1-10, 01:08 PM   #11
 
I just somehow know it is a fact
Quote by Count Iblis View Post
There is a typo in the problem: "79" should be "78".
Sorry mac, it is "79"
 
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