DeltaS=2pir eching

$$\Delta S = 2 \pi r$$

S uchun yechish (complex solution)

$$\left\{\begin{matrix}S=\frac{2\pi r}{\Delta }\text{, }&\Delta \neq 0\\S\in \mathrm{C}\text{, }&r=0\text{ and }\Delta =0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$\Delta S=2\pi r$$

Ikki tarafini \Delta ga bo‘ling.

$$\frac{\Delta S}{\Delta }=\frac{2\pi r}{\Delta }$$

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

$$S=\frac{2\pi r}{\Delta }$$

S uchun yechish

$$\left\{\begin{matrix}S=\frac{2\pi r}{\Delta }\text{, }&\Delta \neq 0\\S\in \mathrm{R}\text{, }&r=0\text{ and }\Delta =0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglama standart shaklda.

$$\Delta S=2\pi r$$

Ikki tarafini \Delta ga bo‘ling.

$$\frac{\Delta S}{\Delta }=\frac{2\pi r}{\Delta }$$

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

$$S=\frac{2\pi r}{\Delta }$$

r uchun yechish

$$r=\frac{S\Delta }{2\pi }$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$2\pi r=\Delta S$$

Tenglama standart shaklda.

$$2\pi r=S\Delta$$

Ikki tarafini 2\pi ga bo‘ling.

$$\frac{2\pi r}{2\pi }=\frac{S\Delta }{2\pi }$$

2\pi ga bo'lish 2\pi ga ko'paytirishni bekor qiladi.

$$r=\frac{S\Delta }{2\pi }$$