x+1/2+3(2x-1)/2geqx+3/4 eching

$$\frac { x + 1 } { 2 } + \frac { 3 ( 2 x - 1 ) } { 2 } \geq \frac { x + 3 } { 4 }$$

x uchun yechish

$$x\geq \frac{7}{13}$$

Yechim qadamlari

Tenglamaning ikkala tarafini 4 ga, 2,4 ning eng kichik karralisiga ko‘paytiring. 4 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$2\left(x+1\right)+2\times 3\left(2x-1\right)\geq x+3$$

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

$$2x+2+2\times 3\left(2x-1\right)\geq x+3$$

6 hosil qilish uchun 2 va 3 ni ko'paytirish.

$$2x+2+6\left(2x-1\right)\geq x+3$$

6 ga 2x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x+2+12x-6\geq x+3$$

14x ni olish uchun 2x va 12x ni birlashtirish.

$$14x+2-6\geq x+3$$

-4 olish uchun 2 dan 6 ni ayirish.

$$14x-4\geq x+3$$

Ikkala tarafdan x ni ayirish.

$$14x-4-x\geq 3$$

13x ni olish uchun 14x va -x ni birlashtirish.

$$13x-4\geq 3$$

4 ni ikki tarafga qo’shing.

$$13x\geq 3+4$$

7 olish uchun 3 va 4'ni qo'shing.

$$13x\geq 7$$

Ikki tarafini 13 ga bo‘ling. 13 musbat bo‘lgani uchun, tengsizlik yo‘nalishi o‘zgarmaydi.

$$x\geq \frac{7}{13}$$