5[2(3-x)-1]leq27 eching

$$5 [ 2 ( 3 - x ) - 1 ] \leq 27$$

x uchun yechish

$$x\geq -\frac{1}{5}$$

Yechim qadamlari

Ikki tarafini 5 ga bo‘ling. 5 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$2\left(3-x\right)-1\leq \frac{27}{5}$$

2 ga 3-x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$6-2x-1\leq \frac{27}{5}$$

5 olish uchun 6 dan 1 ni ayirish.

$$5-2x\leq \frac{27}{5}$$

Ikkala tarafdan 5 ni ayirish.

$$-2x\leq \frac{27}{5}-5$$

5 ni \frac{25}{5} kasrga o‘giring.

$$-2x\leq \frac{27}{5}-\frac{25}{5}$$

\frac{27}{5} va \frac{25}{5} da bir xil maxraji bor, ularning suratini ayirish orqali ayiring.

$$-2x\leq \frac{27-25}{5}$$

2 olish uchun 27 dan 25 ni ayirish.

$$-2x\leq \frac{2}{5}$$

Ikki tarafini -2 ga bo‘ling. -2 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$x\geq \frac{\frac{2}{5}}{-2}$$

\frac{\frac{2}{5}}{-2} ni yagona kasrga aylantiring.

$$x\geq \frac{2}{5\left(-2\right)}$$

-10 hosil qilish uchun 5 va -2 ni ko'paytirish.

$$x\geq \frac{2}{-10}$$

\frac{2}{-10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x\geq -\frac{1}{5}$$