y=alpha+betax+u eching

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

x uchun yechish (complex solution)

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Ikkala tarafdan \alpha ni ayirish.

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

Ikkala tarafdan u ni ayirish.

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

Tenglama standart shaklda.

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

Ikki tarafini \beta ga bo‘ling.

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

\beta ga bo'lish \beta ga ko'paytirishni bekor qiladi.

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

u uchun yechish

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

Yechim qadamlari

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Ikkala tarafdan \alpha ni ayirish.

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

Ikkala tarafdan \beta x ni ayirish.

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

x uchun yechish

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

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

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

Ikkala tarafdan \alpha ni ayirish.

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

Ikkala tarafdan u ni ayirish.

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

Tenglama standart shaklda.

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

Ikki tarafini \beta ga bo‘ling.

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

\beta ga bo'lish \beta ga ko'paytirishni bekor qiladi.

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