Réitigh y=alpha+betax+u

$$y = \alpha + \beta x + u$$

Réitigh do x. (complex solution)

$$\left\{\begin{matrix}x=-\frac{\alpha +u-y}{\beta }\text{, }&\beta \neq 0\\x\in \mathrm{C}\text{, }&y=u+\alpha \text{ and }\beta =0\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

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

$$\alpha +\beta x+u=y$$

Bain \alpha ón dá thaobh.

$$\beta x+u=y-\alpha$$

Bain u ón dá thaobh.

$$\beta x=y-\alpha -u$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\beta x=y-u-\alpha$$

Roinn an dá thaobh faoi \beta .

$$\frac{\beta x}{\beta }=\frac{y-u-\alpha }{\beta }$$

Má roinntear é faoi \beta cuirtear an iolrúchán faoi \beta ar ceal.

$$x=\frac{y-u-\alpha }{\beta }$$

Réitigh do u.

$$u=-x\beta +y-\alpha$$

Céimeanna Réitigh

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

$$\alpha +\beta x+u=y$$

Bain \alpha ón dá thaobh.

$$\beta x+u=y-\alpha$$

Bain \beta x ón dá thaobh.

$$u=y-\alpha -\beta x$$

Réitigh do x.

$$\left\{\begin{matrix}x=-\frac{\alpha +u-y}{\beta }\text{, }&\beta \neq 0\\x\in \mathrm{R}\text{, }&y=u+\alpha \text{ and }\beta =0\end{matrix}\right.$$

Céimeanna chun Cothromóidí Líneacha a Réiteach

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

$$\alpha +\beta x+u=y$$

Bain \alpha ón dá thaobh.

$$\beta x+u=y-\alpha$$

Bain u ón dá thaobh.

$$\beta x=y-\alpha -u$$

Tá an chothromóid i bhfoirm chaighdeánach.

$$\beta x=y-u-\alpha$$

Roinn an dá thaobh faoi \beta .

$$\frac{\beta x}{\beta }=\frac{y-u-\alpha }{\beta }$$

Má roinntear é faoi \beta cuirtear an iolrúchán faoi \beta ar ceal.

$$x=\frac{y-u-\alpha }{\beta }$$