x-1/x-2=1-y/n eching

$$\frac { x - 1 } { x - 2 } = \frac { 1 - y } { n }$$

n uchun yechish

$$\left\{\begin{matrix}n=-\frac{\left(x-2\right)\left(y-1\right)}{x-1}\text{, }&y\neq 1\text{ and }x\neq 2\text{ and }x\neq 1\\n\neq 0\text{, }&y=1\text{ and }x=1\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

n qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini n\left(x-2\right) ga, x-2,n ning eng kichik karralisiga ko‘paytiring.

$$n\left(x-1\right)=\left(x-2\right)\left(1-y\right)$$

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

$$nx-n=\left(x-2\right)\left(1-y\right)$$

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

$$nx-n=x-xy-2+2y$$

n'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(x-1\right)n=x-xy-2+2y$$

Tenglama standart shaklda.

$$\left(x-1\right)n=-xy+x+2y-2$$

Ikki tarafini x-1 ga bo‘ling.

$$\frac{\left(x-1\right)n}{x-1}=\frac{\left(1-y\right)\left(x-2\right)}{x-1}$$

x-1 ga bo'lish x-1 ga ko'paytirishni bekor qiladi.

$$n=\frac{\left(1-y\right)\left(x-2\right)}{x-1}$$

n qiymati 0 teng bo‘lmaydi.

$$n=\frac{\left(1-y\right)\left(x-2\right)}{x-1}\text{, }n\neq 0$$

x uchun yechish

$$x=-\frac{2-n-2y}{y+n-1}$$

$$n\neq 0\text{ and }y\neq 1-n$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

x qiymati 2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini n\left(x-2\right) ga, x-2,n ning eng kichik karralisiga ko‘paytiring.

$$n\left(x-1\right)=\left(x-2\right)\left(1-y\right)$$

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

$$nx-n=\left(x-2\right)\left(1-y\right)$$

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

$$nx-n=x-xy-2+2y$$

Ikkala tarafdan x ni ayirish.

$$nx-n-x=-xy-2+2y$$

xy ni ikki tarafga qo’shing.

$$nx-n-x+xy=-2+2y$$

n ni ikki tarafga qo’shing.

$$nx-x+xy=-2+2y+n$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(n-1+y\right)x=-2+2y+n$$

Tenglama standart shaklda.

$$\left(y+n-1\right)x=2y+n-2$$

Ikki tarafini n-1+y ga bo‘ling.

$$\frac{\left(y+n-1\right)x}{y+n-1}=\frac{2y+n-2}{y+n-1}$$

n-1+y ga bo'lish n-1+y ga ko'paytirishni bekor qiladi.

$$x=\frac{2y+n-2}{y+n-1}$$

x qiymati 2 teng bo‘lmaydi.

$$x=\frac{2y+n-2}{y+n-1}\text{, }x\neq 2$$