y-4=2/3(x+5) eching

$$y - 4 = \frac { 2 } { 3 } ( x + 5 )$$

x uchun yechish

$$x=\frac{3y}{2}-11$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

\frac{2}{3} ga x+5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$y-4=\frac{2}{3}x+\frac{10}{3}$$

Tomonlarni almashtirib, barcha oʻzgaruvchi shartlar chap tomonga oʻtkazing.

$$\frac{2}{3}x+\frac{10}{3}=y-4$$

Ikkala tarafdan \frac{10}{3} ni ayirish.

$$\frac{2}{3}x=y-4-\frac{10}{3}$$

-\frac{22}{3} olish uchun -4 dan \frac{10}{3} ni ayirish.

$$\frac{2}{3}x=y-\frac{22}{3}$$

Tenglamaning ikki tarafini \frac{2}{3} ga bo'lish, bu kasrni qaytarish orqali ikkala tarafga ko'paytirish bilan aynidir.

$$\frac{\frac{2}{3}x}{\frac{2}{3}}=\frac{y-\frac{22}{3}}{\frac{2}{3}}$$

\frac{2}{3} ga bo'lish \frac{2}{3} ga ko'paytirishni bekor qiladi.

$$x=\frac{y-\frac{22}{3}}{\frac{2}{3}}$$

y-\frac{22}{3} ni \frac{2}{3} ga bo'lish y-\frac{22}{3} ga k'paytirish \frac{2}{3} ga qaytarish.

$$x=\frac{3y}{2}-11$$

y uchun yechish

$$y=\frac{2\left(x+11\right)}{3}$$

Yechim qadamlari

\frac{2}{3} ga x+5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$y-4=\frac{2}{3}x+\frac{10}{3}$$

4 ni ikki tarafga qo’shing.

$$y=\frac{2}{3}x+\frac{10}{3}+4$$

\frac{22}{3} olish uchun \frac{10}{3} va 4'ni qo'shing.

$$y=\frac{2}{3}x+\frac{22}{3}$$