3x-4/5+1/8leq6x+7/8-x+1 eching

$$\frac{ 3x-4 }{ 5 } + \frac{ 1 }{ 8 } \leq \frac{ 6x+7 }{ 8 } -x+1$$

x uchun yechish

$$x\leq 3$$

Yechim qadamlari

Tenglamaning ikkala tarafini 40 ga, 5,8 ning eng kichik karralisiga ko‘paytiring. 40 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$8\left(3x-4\right)+5\leq 5\left(6x+7\right)-40x+40$$

8 ga 3x-4 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$24x-32+5\leq 5\left(6x+7\right)-40x+40$$

-27 olish uchun -32 va 5'ni qo'shing.

$$24x-27\leq 5\left(6x+7\right)-40x+40$$

5 ga 6x+7 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$24x-27\leq 30x+35-40x+40$$

-10x ni olish uchun 30x va -40x ni birlashtirish.

$$24x-27\leq -10x+35+40$$

75 olish uchun 35 va 40'ni qo'shing.

$$24x-27\leq -10x+75$$

10x ni ikki tarafga qo’shing.

$$24x-27+10x\leq 75$$

34x ni olish uchun 24x va 10x ni birlashtirish.

$$34x-27\leq 75$$

27 ni ikki tarafga qo’shing.

$$34x\leq 75+27$$

102 olish uchun 75 va 27'ni qo'shing.

$$34x\leq 102$$

Ikki tarafini 34 ga bo‘ling. 34 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$x\leq \frac{102}{34}$$

3 ni olish uchun 102 ni 34 ga bo‘ling.

$$x\leq 3$$