9-3(x+2)geq8x-5 eching

$$9-3(x+2) \geq 8x-5$$

x uchun yechish

$$x\leq \frac{8}{11}$$

Yechim qadamlari

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

$$9-3x-6\geq 8x-5$$

3 olish uchun 9 dan 6 ni ayirish.

$$3-3x\geq 8x-5$$

Ikkala tarafdan 8x ni ayirish.

$$3-3x-8x\geq -5$$

-11x ni olish uchun -3x va -8x ni birlashtirish.

$$3-11x\geq -5$$

Ikkala tarafdan 3 ni ayirish.

$$-11x\geq -5-3$$

-8 olish uchun -5 dan 3 ni ayirish.

$$-11x\geq -8$$

Ikki tarafini -11 ga bo‘ling. -11 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$x\leq \frac{-8}{-11}$$

Ikkala surat va maxrajdan manfiy belgini olib tashlash bilan \frac{-8}{-11} kasrini \frac{8}{11} ga soddalashtirish mumkin.

$$x\leq \frac{8}{11}$$