bx+c/3a=y^2 eching

$$\frac { b x + c } { 3 a } = y ^ { 2 }$$

b uchun yechish (complex solution)

$$\left\{\begin{matrix}b=-\frac{c-3ay^{2}}{x}\text{, }&x\neq 0\text{ and }a\neq 0\\b\in \mathrm{C}\text{, }&c=3ay^{2}\text{ and }x=0\text{ and }a\neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini 3a ga ko'paytirish.

$$bx+c=3ay^{2}$$

Ikkala tarafdan c ni ayirish.

$$bx=3ay^{2}-c$$

Tenglama standart shaklda.

$$xb=3ay^{2}-c$$

Ikki tarafini x ga bo‘ling.

$$\frac{xb}{x}=\frac{3ay^{2}-c}{x}$$

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

$$b=\frac{3ay^{2}-c}{x}$$

a uchun yechish

$$\left\{\begin{matrix}a=\frac{bx+c}{3y^{2}}\text{, }&\left(c\neq 0\text{ or }x\neq 0\right)\text{ and }\left(x=0\text{ or }b\neq -\frac{c}{x}\right)\text{ and }\left(b\neq 0\text{ or }c\neq 0\right)\text{ and }y\neq 0\text{ and }c\neq -bx\\a\neq 0\text{, }&y=0\text{ and }c=-bx\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

a qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 3a ga ko'paytirish.

$$bx+c=3ay^{2}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$3ay^{2}=bx+c$$

Tenglama standart shaklda.

$$3y^{2}a=bx+c$$

Ikki tarafini 3y^{2} ga bo‘ling.

$$\frac{3y^{2}a}{3y^{2}}=\frac{bx+c}{3y^{2}}$$

3y^{2} ga bo'lish 3y^{2} ga ko'paytirishni bekor qiladi.

$$a=\frac{bx+c}{3y^{2}}$$

a qiymati 0 teng bo‘lmaydi.

$$a=\frac{bx+c}{3y^{2}}\text{, }a\neq 0$$

b uchun yechish

$$\left\{\begin{matrix}b=-\frac{c-3ay^{2}}{x}\text{, }&x\neq 0\text{ and }a\neq 0\\b\in \mathrm{R}\text{, }&c=3ay^{2}\text{ and }x=0\text{ and }a\neq 0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini 3a ga ko'paytirish.

$$bx+c=3ay^{2}$$

Ikkala tarafdan c ni ayirish.

$$bx=3ay^{2}-c$$

Tenglama standart shaklda.

$$xb=3ay^{2}-c$$

Ikki tarafini x ga bo‘ling.

$$\frac{xb}{x}=\frac{3ay^{2}-c}{x}$$

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

$$b=\frac{3ay^{2}-c}{x}$$