2[1-(4x-3)+2(x+6)]=3[x+4(2x-5)-6(3x+2)] eching

$$2 [ 1 - ( 4 x - 3 ) + 2 ( x + 6 ) ] = 3 [ x + 4 ( 2 x - 5 ) - 6 ( 3 x + 2 ) ]$$

x uchun yechish

$$x = -\frac{128}{23} = -5\frac{13}{23} \approx -5,565217391$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

4x-3 teskarisini topish uchun har birining teskarisini toping.

$$2\left(1-4x-\left(-3\right)+2\left(x+6\right)\right)=3\left(x+4\left(2x-5\right)-6\left(3x+2\right)\right)$$

-3 ning teskarisi 3 ga teng.

$$2\left(1-4x+3+2\left(x+6\right)\right)=3\left(x+4\left(2x-5\right)-6\left(3x+2\right)\right)$$

4 olish uchun 1 va 3'ni qo'shing.

$$2\left(4-4x+2\left(x+6\right)\right)=3\left(x+4\left(2x-5\right)-6\left(3x+2\right)\right)$$

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

$$2\left(4-4x+2x+12\right)=3\left(x+4\left(2x-5\right)-6\left(3x+2\right)\right)$$

-2x ni olish uchun -4x va 2x ni birlashtirish.

$$2\left(4-2x+12\right)=3\left(x+4\left(2x-5\right)-6\left(3x+2\right)\right)$$

16 olish uchun 4 va 12'ni qo'shing.

$$2\left(16-2x\right)=3\left(x+4\left(2x-5\right)-6\left(3x+2\right)\right)$$

2 ga 16-2x ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$32-4x=3\left(x+4\left(2x-5\right)-6\left(3x+2\right)\right)$$

4 ga 2x-5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$32-4x=3\left(x+8x-20-6\left(3x+2\right)\right)$$

9x ni olish uchun x va 8x ni birlashtirish.

$$32-4x=3\left(9x-20-6\left(3x+2\right)\right)$$

-6 ga 3x+2 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$32-4x=3\left(9x-20-18x-12\right)$$

-9x ni olish uchun 9x va -18x ni birlashtirish.

$$32-4x=3\left(-9x-20-12\right)$$

-32 olish uchun -20 dan 12 ni ayirish.

$$32-4x=3\left(-9x-32\right)$$

3 ga -9x-32 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$32-4x=-27x-96$$

27x ni ikki tarafga qo’shing.

$$32-4x+27x=-96$$

23x ni olish uchun -4x va 27x ni birlashtirish.

$$32+23x=-96$$

Ikkala tarafdan 32 ni ayirish.

$$23x=-96-32$$

-128 olish uchun -96 dan 32 ni ayirish.

$$23x=-128$$

Ikki tarafini 23 ga bo‘ling.

$$x=\frac{-128}{23}$$

\frac{-128}{23} kasri manfiy belgini olib tashlash bilan -\frac{128}{23} sifatida qayta yozilishi mumkin.

$$x=-\frac{128}{23}$$