varepsilon=pi/E(sigma_{1}-v(sigma_{2}+sigma_{3})) eching

$$\varepsilon = \frac { \pi } { E } ( \sigma _ { 1 } - v ( \sigma _ { 2 } + \sigma _ { 3 } ) )$$

E uchun yechish

$$\left\{\begin{matrix}E=\frac{\pi \left(\sigma _{1}-v\sigma _{3}-v\sigma _{2}\right)}{\epsilon }\text{, }&\sigma _{1}\neq v\left(\sigma _{2}+\sigma _{3}\right)\text{ and }\epsilon \neq 0\text{ and }\sigma _{1}\neq v\sigma _{2}+v\sigma _{3}\\E\neq 0\text{, }&\epsilon =0\text{ and }\sigma _{1}=v\left(\sigma _{2}+\sigma _{3}\right)\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

E qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini E ga ko'paytirish.

$$\epsilon E=\pi \left(\sigma _{1}-v\left(\sigma _{2}+\sigma _{3}\right)\right)$$

v ga \sigma _{2}+\sigma _{3} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\epsilon E=\pi \left(\sigma _{1}-\left(v\sigma _{2}+v\sigma _{3}\right)\right)$$

v\sigma _{2}+v\sigma _{3} teskarisini topish uchun har birining teskarisini toping.

$$\epsilon E=\pi \left(\sigma _{1}-v\sigma _{2}-v\sigma _{3}\right)$$

\pi ga \sigma _{1}-v\sigma _{2}-v\sigma _{3} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\epsilon E=\pi \sigma _{1}-\pi v\sigma _{2}-\pi v\sigma _{3}$$

Tenglama standart shaklda.

$$\epsilon E=\pi \sigma _{1}-\pi v\sigma _{3}-\pi v\sigma _{2}$$

Ikki tarafini \epsilon ga bo‘ling.

$$\frac{\epsilon E}{\epsilon }=\frac{\pi \left(\sigma _{1}-v\sigma _{3}-v\sigma _{2}\right)}{\epsilon }$$

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

$$E=\frac{\pi \left(\sigma _{1}-v\sigma _{3}-v\sigma _{2}\right)}{\epsilon }$$

E qiymati 0 teng bo‘lmaydi.

$$E=\frac{\pi \left(\sigma _{1}-v\sigma _{3}-v\sigma _{2}\right)}{\epsilon }\text{, }E\neq 0$$

v uchun yechish

$$\left\{\begin{matrix}v=\frac{\pi \sigma _{1}-E\epsilon }{\pi \left(\sigma _{2}+\sigma _{3}\right)}\text{, }&E\neq 0\text{ and }\sigma _{2}\neq -\sigma _{3}\\v\in \mathrm{R}\text{, }&\sigma _{1}=\frac{E\epsilon }{\pi }\text{ and }\sigma _{2}=-\sigma _{3}\text{ and }E\neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini E ga ko'paytirish.

$$\epsilon E=\pi \left(\sigma _{1}-v\left(\sigma _{2}+\sigma _{3}\right)\right)$$

v ga \sigma _{2}+\sigma _{3} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\epsilon E=\pi \left(\sigma _{1}-\left(v\sigma _{2}+v\sigma _{3}\right)\right)$$

v\sigma _{2}+v\sigma _{3} teskarisini topish uchun har birining teskarisini toping.

$$\epsilon E=\pi \left(\sigma _{1}-v\sigma _{2}-v\sigma _{3}\right)$$

\pi ga \sigma _{1}-v\sigma _{2}-v\sigma _{3} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$\epsilon E=\pi \sigma _{1}-\pi v\sigma _{2}-\pi v\sigma _{3}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\pi \sigma _{1}-\pi v\sigma _{2}-\pi v\sigma _{3}=\epsilon E$$

Ikkala tarafdan \pi \sigma _{1} ni ayirish.

$$-\pi v\sigma _{2}-\pi v\sigma _{3}=\epsilon E-\pi \sigma _{1}$$

Shartlarni qayta saralash.

$$-\pi v\sigma _{2}-\pi v\sigma _{3}=E\epsilon -\pi \sigma _{1}$$

v'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(-\pi \sigma _{2}-\pi \sigma _{3}\right)v=E\epsilon -\pi \sigma _{1}$$

Ikki tarafini -\pi \sigma _{2}-\pi \sigma _{3} ga bo‘ling.

$$\frac{\left(-\pi \sigma _{2}-\pi \sigma _{3}\right)v}{-\pi \sigma _{2}-\pi \sigma _{3}}=\frac{E\epsilon -\pi \sigma _{1}}{-\pi \sigma _{2}-\pi \sigma _{3}}$$

-\pi \sigma _{2}-\pi \sigma _{3} ga bo'lish -\pi \sigma _{2}-\pi \sigma _{3} ga ko'paytirishni bekor qiladi.

$$v=\frac{E\epsilon -\pi \sigma _{1}}{-\pi \sigma _{2}-\pi \sigma _{3}}$$

\epsilon E-\pi \sigma _{1} ni -\pi \sigma _{2}-\pi \sigma _{3} ga bo'lish.

$$v=\frac{E\epsilon -\pi \sigma _{1}}{-\pi \left(\sigma _{2}+\sigma _{3}\right)}$$