(x-1)x^2+(x+4)y+x+7=0 eching

$$(x-1) { x }^{ 2 } +(x+4)y+x+7=0$$

y uchun yechish

$$y=-\frac{x^{3}-x^{2}+x+7}{x+4}$$

$$x\neq -4$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

x-1 ga x^{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{3}-x^{2}+\left(x+4\right)y+x+7=0$$

x+4 ga y ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x^{3}-x^{2}+xy+4y+x+7=0$$

Ikkala tarafdan x^{3} ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$-x^{2}+xy+4y+x+7=-x^{3}$$

x^{2} ni ikki tarafga qo’shing.

$$xy+4y+x+7=-x^{3}+x^{2}$$

Ikkala tarafdan x ni ayirish.

$$xy+4y+7=-x^{3}+x^{2}-x$$

Ikkala tarafdan 7 ni ayirish.

$$xy+4y=-x^{3}+x^{2}-x-7$$

y'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x+4\right)y=-x^{3}+x^{2}-x-7$$

Ikki tarafini x+4 ga bo‘ling.

$$\frac{\left(x+4\right)y}{x+4}=\frac{-x^{3}+x^{2}-x-7}{x+4}$$

x+4 ga bo'lish x+4 ga ko'paytirishni bekor qiladi.

$$y=\frac{-x^{3}+x^{2}-x-7}{x+4}$$