Pagina:Werkecarlf03gausrich.djvu/354

5.

E combinatione aequationum [1] et [2] facile derivantur sequentes:

${\begin{aligned}&-\alpha aa+\gamma aA=\alpha G\\&-\beta bb+\gamma bB=\beta G\\&\gamma (AA+BB+CC)-\alpha aA-\beta bB=\gamma G\end{aligned}}$

unde fit porro

$\alpha ={\frac {\gamma aA}{aa+G}}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$	[3]
$\beta ={\frac {\gamma bB}{bb+G}}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$	[4]
$AA+BB+CC-{\frac {aaAA}{aa+G}}-{\frac {bbB}{bb+G}}=G$

Ultimam sic quoque exhibere possumus

${\frac {AA}{aa+G}}+{\frac {BB}{bb+G}}+{\frac {CC}{G}}=1\ldots \ldots \ldots \ldots \,$[5]

Perinde e combinatione aequationum [1] et [2] deducimus

${\begin{aligned}&\alpha ^{\prime }aa-\gamma ^{\prime }aA=\alpha ^{\prime }G^{\prime }\\&\beta ^{\prime }bb-\gamma ^{\prime }bB=\beta ^{\prime }G^{\prime }\\&-\gamma ^{\prime }(AA+BB+CC)+\alpha ^{\prime }aA+\beta ^{\prime }bB=\gamma ^{\prime }G^{\prime }\end{aligned}}$

atque hinc

$\alpha ^{\prime }={\frac {\gamma ^{\prime }aA}{aa-G^{\prime }}}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ..$	[6]
$\beta ^{\prime }={\frac {\gamma ^{\prime }bB}{bb-G^{\prime }}}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$	[7]
${\frac {AA}{aa-G^{\prime }}}+{\frac {BB}{bb-G^{\prime }}}-{\frac {CC}{G^{\prime }}}=1\ldots \ldots \ldots \ldots .$	[8]

et prorsus simili modo

$\alpha ^{\prime \prime }={\frac {\gamma ^{\prime \prime }aA}{aa-G^{\prime \prime }}}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$	[9]
$\beta ^{\prime \prime }={\frac {\gamma ^{\prime \prime }bB}{bb-G^{\prime \prime }}}\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$	[10]
${\frac {AA}{aa-G^{\prime \prime }}}+{\frac {BB}{bb-G^{\prime \prime }}}-{\frac {CC}{G^{\prime \prime }}}=1\ldots \ldots \ldots \ldots .$	[11]

Patet itaque, $G,$ $-G^{\prime },$ $-G^{\prime \prime }$ esse radices aequationis

${\frac {AA}{aa+x}}+{\frac {BB}{bb+x}}+{\frac {CC}{x}}=1$