Réitigh m+5n=m-n/12m

$$m + 5 n = \frac { m - n } { 12 m }$$

Réitigh do n.

$$n=-\frac{m\left(12m-1\right)}{60m+1}$$

$$m\neq -\frac{1}{60}\text{ and }m\neq 0$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

Méadaigh an dá thaobh den chothromóid faoi 12m.

$$12mm+5n\times 12m=m-n$$

Méadaigh m agus m chun m^{2} a fháil.

$$12m^{2}+5n\times 12m=m-n$$

Méadaigh 5 agus 12 chun 60 a fháil.

$$12m^{2}+60nm=m-n$$

Cuir n leis an dá thaobh.

$$12m^{2}+60nm+n=m$$

Bain 12m^{2} ón dá thaobh.

$$60nm+n=m-12m^{2}$$

Comhcheangail na téarmaí ar fad ina bhfuil n.

$$\left(60m+1\right)n=m-12m^{2}$$

Roinn an dá thaobh faoi 60m+1.

$$\frac{\left(60m+1\right)n}{60m+1}=\frac{m\left(1-12m\right)}{60m+1}$$

Má roinntear é faoi 60m+1 cuirtear an iolrúchán faoi 60m+1 ar ceal.

$$n=\frac{m\left(1-12m\right)}{60m+1}$$

Réitigh do m. (complex solution)

$$\left\{\begin{matrix}\\m=\frac{\sqrt{3600n^{2}-168n+1}}{24}-\frac{5n}{2}+\frac{1}{24}\text{, }&\text{unconditionally}\\m=-\frac{\sqrt{3600n^{2}-168n+1}}{24}-\frac{5n}{2}+\frac{1}{24}\text{, }&n\neq 0\end{matrix}\right.$$

Réitigh do m.

$$\left\{\begin{matrix}m=-\frac{\sqrt{3600n^{2}-168n+1}}{24}-\frac{5n}{2}+\frac{1}{24}\text{, }&n\geq \frac{\sqrt{6}}{150}+\frac{7}{300}\text{ or }\left(n\neq 0\text{ and }n\leq -\frac{\sqrt{6}}{150}+\frac{7}{300}\right)\\m=\frac{\sqrt{3600n^{2}-168n+1}}{24}-\frac{5n}{2}+\frac{1}{24}\text{, }&n\geq \frac{\sqrt{6}}{150}+\frac{7}{300}\text{ or }n\leq -\frac{\sqrt{6}}{150}+\frac{7}{300}\end{matrix}\right.$$