A=-5(x+2)quadB eching

$$A = - 5 ( x + 2 ) \quad B$$

B uchun yechish (complex solution)

$$\left\{\begin{matrix}B=-\frac{A}{5\left(x+2\right)}\text{, }&x\neq -2\\B\in \mathrm{C}\text{, }&A=0\text{ and }x=-2\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

-5 ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$A=\left(-5x-10\right)B$$

-5x-10 ga B ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$A=-5xB-10B$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-5xB-10B=A$$

B'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(-5x-10\right)B=A$$

Ikki tarafini -5x-10 ga bo‘ling.

$$\frac{\left(-5x-10\right)B}{-5x-10}=\frac{A}{-5x-10}$$

-5x-10 ga bo'lish -5x-10 ga ko'paytirishni bekor qiladi.

$$B=\frac{A}{-5x-10}$$

A ni -5x-10 ga bo'lish.

$$B=-\frac{A}{5\left(x+2\right)}$$

A uchun yechish

$$A=-5B\left(x+2\right)$$

Yechim qadamlari

-5 ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$A=\left(-5x-10\right)B$$

-5x-10 ga B ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$A=-5xB-10B$$

B uchun yechish

$$\left\{\begin{matrix}B=-\frac{A}{5\left(x+2\right)}\text{, }&x\neq -2\\B\in \mathrm{R}\text{, }&A=0\text{ and }x=-2\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

-5 ga x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$A=\left(-5x-10\right)B$$

-5x-10 ga B ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$A=-5xB-10B$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$-5xB-10B=A$$

B'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(-5x-10\right)B=A$$

Ikki tarafini -5x-10 ga bo‘ling.

$$\frac{\left(-5x-10\right)B}{-5x-10}=\frac{A}{-5x-10}$$

-5x-10 ga bo'lish -5x-10 ga ko'paytirishni bekor qiladi.

$$B=\frac{A}{-5x-10}$$

A ni -5x-10 ga bo'lish.

$$B=-\frac{A}{5\left(x+2\right)}$$