How to solve this system of equations of second degree in set of real numbers

$$15(x+y)=(15+x+y)x$$ $$(15+x+y)x=12\sqrt{x^2-6^2}+9\sqrt{y^2-(9/2)^2}+108 $$

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    $\begingroup$ Just out of curiosity, how do you obtain this system ? (maybe there is some simpler way to solve your initial problem?) $\endgroup$ – Jean-Claude Arbaut Mar 25 '13 at 13:12
  • $\begingroup$ "The question is widely applicable to a large audience" says the bounty message. Is it really? If there's a general method being sought here, I don't see it explicitly asked for. $\endgroup$ – Ben Millwood Apr 3 '13 at 11:09
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    $\begingroup$ May be the following remark will be useful: the right part of the 2nd equation can be rewritten as $$(\sqrt{x^2-6^2}+6)^2 + (\sqrt{y^2-(9/2)^2}+9/2)^2+ 108-x^2-y^2$$ $\endgroup$ – Boris Novikov Apr 4 '13 at 10:17

The first thing that comes to mind is to use the first equation to express one variable using the other: $$15(x+y)=(15+x+y)x\iff 15y=x^2+xy\iff y(15-x)=x^2\iff y=\frac{x^2}{15-x}$$ (since $x=15$ is not a solution). Now substitute this into the second equation: $$15\left(x+\frac{x^2}{15-x}\right)=\frac{225x}{15-x}=12\sqrt{x^2-6^2}+9\sqrt{\frac{x^4}{(15-x)^2}-(9/2)^2}+108$$ $$\begin{align*}\frac{75x}{15-x}-36&=4\sqrt{x^2-6^2}+3\sqrt{\frac{x^4}{(15-x)^2}-(9/2)^2}\\ 3\frac{37x-180}{15-x}&=4\sqrt{x^2-6^2}+3\frac{\sqrt{4x^4-81(15-x)^2}}{2(15-x)} \end{align*}$$ Squaring: $$9\frac{(37x-180)^2}{(15-x)^2}=16(x^2-6^2)+3\frac{4x^4-81(15-x)^2}{4(15-x)^2}+24\frac{\sqrt{x^2-6^2}\sqrt{4x^4-81(15-x)^2}}{2(15-x)}$$ Rearanging: $$\frac{-76 x^4+1920 x^3+37431 x^2-555930 x+1739475}{12(15-x)}=\sqrt{x^2-6^2}\sqrt{4x^4-81(15-x)^2}$$ Squaring and rearanging (using WA) once again, we have: $$5200 x^8-274560 x^7-2100312 x^6+226914480 x^5-978099543 x^4-35070717420 x^3+439302361950 x^2-1928383969500 x+3004515635625=0$$ Which has 4 real and 4 non-real roots. Following the above calculation, one can check that the conditions before each squaring imply that none of those solutions solves the initial equation. Hence there is no solution.

  • $\begingroup$ It is a little cleaner to replace the left side of the second with $15(x+y)$, but will still be a mess. $\endgroup$ – Ross Millikan Mar 25 '13 at 13:29
  • $\begingroup$ After substitution and simplification, and if I made no mistake, I get an 8th degree polynomial in x with 4 real roots, none of which gives a solution of the initial system. Thus no solution. $\endgroup$ – Jean-Claude Arbaut Mar 25 '13 at 13:33
  • $\begingroup$ Can you prove that $\endgroup$ – Milingona Ana Mar 25 '13 at 13:41
  • $\begingroup$ With Maxima, yes. By hand it's a bit harder :-) $\endgroup$ – Jean-Claude Arbaut Mar 25 '13 at 13:48
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    $\begingroup$ In your line after "Squaring:", haven't you forgotten to square the 3? $\endgroup$ – Ben Millwood Apr 3 '13 at 11:08

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