F+2=H-10k(E+2) eching

$$F + 2 = H - 10 k ( E + 2 )$$

E uchun yechish

$$\left\{\begin{matrix}E=\frac{-F+H-20k-2}{10k}\text{, }&k\neq 0\\E\in \mathrm{R}\text{, }&F=H-2\text{ and }k=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$H-10k\left(E+2\right)=F+2$$

-10k ga E+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$H-10kE-20k=F+2$$

Ikkala tarafdan H ni ayirish.

$$-10kE-20k=F+2-H$$

20k ni ikki tarafga qo’shing.

$$-10kE=F+2-H+20k$$

Tenglama standart shaklda.

$$\left(-10k\right)E=F-H+20k+2$$

Ikki tarafini -10k ga bo‘ling.

$$\frac{\left(-10k\right)E}{-10k}=\frac{F-H+20k+2}{-10k}$$

-10k ga bo'lish -10k ga ko'paytirishni bekor qiladi.

$$E=\frac{F-H+20k+2}{-10k}$$

F-H+2+20k ni -10k ga bo'lish.

$$E=-\frac{F-H+20k+2}{10k}$$

F uchun yechish

$$F=-10Ek+H-20k-2$$

Yechim qadamlari

Ikkala tarafdan 2 ni ayirish.

$$F=H-10k\left(E+2\right)-2$$

-10k ga E+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$F=H-10kE-20k-2$$