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Solving 2 equation with 2 variables (hard)

by Zetison
Tags: equation, solving, variables
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Zetison
#1
Apr1-10, 06:00 AM
P: 35


I really need some help here (no program is allowed )
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Pengwuino
#2
Apr1-10, 06:02 AM
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What work have you done on the problem so far?
HallsofIvy
#3
Apr1-10, 07:15 AM
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Thanks
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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.

Zetison
#4
Apr1-10, 07:40 AM
P: 35
Solving 2 equation with 2 variables (hard)

The problem is to find exact values, not estimatet values. Therefore: do not use any programs...
Gib Z
#5
Apr1-10, 07:46 AM
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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.
Zetison
#6
Apr1-10, 07:57 AM
P: 35
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...
Gib Z
#7
Apr1-10, 08:09 AM
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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?
Zetison
#8
Apr1-10, 08:18 AM
P: 35
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.
Count Iblis
#9
Apr1-10, 10:16 AM
P: 2,158
There is a typo in the problem: "79" should be "78".
Gib Z
#10
Apr1-10, 10:48 AM
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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 ?
Zetison
#11
Apr1-10, 01:08 PM
P: 35
I just somehow know it is a fact
Quote 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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