Réitigh 100=F+F1.618+F2.618

$$100 = F+F1.618+F2.618$$

Réitigh do F.

$$F=-\frac{309F_{1}}{500}-\frac{309F_{2}}{500}+100$$

Céimeanna Réitigh

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$F+F_{1}\times 0.618+F_{2}\times 0.618=100$$

Bain F_{1}\times 0.618 ón dá thaobh.

$$F+F_{2}\times 0.618=100-F_{1}\times 0.618$$

Bain F_{2}\times 0.618 ón dá thaobh.

$$F=100-F_{1}\times 0.618-F_{2}\times 0.618$$

Méadaigh -1 agus 0.618 chun -0.618 a fháil.

$$F=100-0.618F_{1}-F_{2}\times 0.618$$

Méadaigh -1 agus 0.618 chun -0.618 a fháil.

$$F=100-0.618F_{1}-0.618F_{2}$$

Réitigh do F_1.

$$F_{1}=-\frac{500F}{309}-F_{2}+\frac{50000}{309}$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Athraigh na taobhanna ionas go mbeidh na téarmaí inathraitheacha ar fad ar an taobh clé.

$$F+F_{1}\times 0.618+F_{2}\times 0.618=100$$

Bain F ón dá thaobh.

$$F_{1}\times 0.618+F_{2}\times 0.618=100-F$$

Bain F_{2}\times 0.618 ón dá thaobh.

$$F_{1}\times 0.618=100-F-F_{2}\times 0.618$$

Méadaigh -1 agus 0.618 chun -0.618 a fháil.

$$F_{1}\times 0.618=100-F-0.618F_{2}$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$0.618F_{1}=-\frac{309F_{2}}{500}-F+100$$

Roinn an dá thaobh den chothromóid faoi 0.618, arb ionann é sin agus an dá thaobh a mhéadú faoi dheilín an chodáin.

$$\frac{0.618F_{1}}{0.618}=\frac{-\frac{309F_{2}}{500}-F+100}{0.618}$$

Má roinntear é faoi 0.618 cuirtear an iolrúchán faoi 0.618 ar ceal.

$$F_{1}=\frac{-\frac{309F_{2}}{500}-F+100}{0.618}$$

Roinn 100-F-\frac{309F_{2}}{500} faoi 0.618 trí 100-F-\frac{309F_{2}}{500} a mhéadú faoi dheilín 0.618.

$$F_{1}=-\frac{500F}{309}-F_{2}+\frac{50000}{309}$$