x+2/3leq5x+1/2 eching

$$\frac { x + 2 } { 3 } \leq \frac { 5 x + 1 } { 2 }$$

x uchun yechish

$$x\geq \frac{1}{13}$$

Yechim qadamlari

Tenglamaning ikkala tarafini 6 ga, 3,2 ning eng kichik karralisiga ko‘paytiring. 6 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$2\left(x+2\right)\leq 3\left(5x+1\right)$$

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

$$2x+4\leq 3\left(5x+1\right)$$

3 ga 5x+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x+4\leq 15x+3$$

Ikkala tarafdan 15x ni ayirish.

$$2x+4-15x\leq 3$$

-13x ni olish uchun 2x va -15x ni birlashtirish.

$$-13x+4\leq 3$$

Ikkala tarafdan 4 ni ayirish.

$$-13x\leq 3-4$$

-1 olish uchun 3 dan 4 ni ayirish.

$$-13x\leq -1$$

Ikki tarafini -13 ga bo‘ling. -13 manfiy boʻlgani uchun tengsizlikning yo‘nalishi o‘zgaradi.

$$x\geq \frac{-1}{-13}$$

Ikkala surat va maxrajdan manfiy belgini olib tashlash bilan \frac{-1}{-13} kasrini \frac{1}{13} ga soddalashtirish mumkin.

$$x\geq \frac{1}{13}$$