f(x)=x^2+5x+6quadyquadg(x)=x^2-3x-10 eching

$$f ( x ) = x ^ { 2 } + 5 x + 6 \quad y \quad g ( x ) = x ^ { 2 } - 3 x - 10$$

g uchun yechish (complex solution)

$$\left\{\begin{matrix}g=-\frac{4x+5}{3xy}\text{, }&y\neq 0\text{ and }x\neq 0\\g\in \mathrm{C}\text{, }&x=-\frac{5}{4}\text{ and }y=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Ikkala tarafdan x^{2} ni ayirish.

$$5x+6ygx=x^{2}-3x-10-x^{2}$$

0 ni olish uchun x^{2} va -x^{2} ni birlashtirish.

$$5x+6ygx=-3x-10$$

Ikkala tarafdan 5x ni ayirish.

$$6ygx=-3x-10-5x$$

-8x ni olish uchun -3x va -5x ni birlashtirish.

$$6ygx=-8x-10$$

Tenglama standart shaklda.

$$6xyg=-8x-10$$

Ikki tarafini 6yx ga bo‘ling.

$$\frac{6xyg}{6xy}=\frac{-8x-10}{6xy}$$

6yx ga bo'lish 6yx ga ko'paytirishni bekor qiladi.

$$g=\frac{-8x-10}{6xy}$$

-10-8x ni 6yx ga bo'lish.

$$g=-\frac{4x+5}{3xy}$$

g uchun yechish

$$\left\{\begin{matrix}g=-\frac{4x+5}{3xy}\text{, }&y\neq 0\text{ and }x\neq 0\\g\in \mathrm{R}\text{, }&x=-\frac{5}{4}\text{ and }y=0\end{matrix}\right,$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Ikkala tarafdan x^{2} ni ayirish.

$$5x+6ygx=x^{2}-3x-10-x^{2}$$

0 ni olish uchun x^{2} va -x^{2} ni birlashtirish.

$$5x+6ygx=-3x-10$$

Ikkala tarafdan 5x ni ayirish.

$$6ygx=-3x-10-5x$$

-8x ni olish uchun -3x va -5x ni birlashtirish.

$$6ygx=-8x-10$$

Tenglama standart shaklda.

$$6xyg=-8x-10$$

Ikki tarafini 6yx ga bo‘ling.

$$\frac{6xyg}{6xy}=\frac{-8x-10}{6xy}$$

6yx ga bo'lish 6yx ga ko'paytirishni bekor qiladi.

$$g=\frac{-8x-10}{6xy}$$

-8x-10 ni 6yx ga bo'lish.

$$g=-\frac{4x+5}{3xy}$$

x uchun yechish

$$x=-\frac{5}{3gy+4}$$

$$g=0\text{ or }y\neq -\frac{4}{3g}$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Ikkala tarafdan x^{2} ni ayirish.

$$x^{2}+5x+6ygx-x^{2}=-3x-10$$

0 ni olish uchun x^{2} va -x^{2} ni birlashtirish.

$$5x+6ygx=-3x-10$$

3x ni ikki tarafga qo’shing.

$$5x+6ygx+3x=-10$$

8x ni olish uchun 5x va 3x ni birlashtirish.

$$8x+6ygx=-10$$

x'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(8+6yg\right)x=-10$$

Tenglama standart shaklda.

$$\left(6gy+8\right)x=-10$$

Ikki tarafini 6gy+8 ga bo‘ling.

$$\frac{\left(6gy+8\right)x}{6gy+8}=-\frac{10}{6gy+8}$$

6gy+8 ga bo'lish 6gy+8 ga ko'paytirishni bekor qiladi.

$$x=-\frac{10}{6gy+8}$$

-10 ni 6gy+8 ga bo'lish.

$$x=-\frac{5}{3gy+4}$$