Solving Nonlinear System: 3x2+2y2=35, 4x2-3y2=24

AI Thread Summary
The discussion focuses on solving the nonlinear system of equations 3x² + 2y² = 35 and 4x² - 3y² = 24. The original poster struggles with rearranging the equations and arrives at incorrect values for x. Suggestions include multiplying the first equation by 3 and the second by 2 to simplify the addition of the equations. A key point raised is the arithmetic error in the rearrangement, particularly in the calculation of y². The correct solutions to the system are identified as (–3, –2), (–3, 2), (3, –2), and (3, 2).
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Homework Statement


solve the system of 3x^{2}+2y^{2}=35 and
4x^{2}-3y^{2}=24


Homework Equations





The Attempt at a Solution


I re arranged for y^2 and got 1\frac{1}{3}x^{2}-16=y^{2} I keep getting x to equal \pm 2.473 this is clearly wrong, the answers should be (–3, –2), (–3, 2), (3, –2), and (3, 2). What am I doing wrong?
 
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Instead of messing with fractions, why not just multiply the first equation by 3, the second equation by 2 and then just add them?
 
rock.freak667 said:
Instead of messing with fractions, why not just multiply the first equation by 3, the second equation by 2 and then just add them?

Thanks for the help but I still have a question, after I combined the two equations how come it worked and it didn’t work when I rearranged them. Did I mess up?
 
Looks like a basic arithmetic error. Because of the "1\frac{1}{3}", which would better be left 4/3, it looks like you solved the second equation for y2: 3y^2= 4x^2- 24 so y^2= (4/3)x^2- 8. 24/3= 8, not 16.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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