Question

Cho \(2\left(x-3\right)=3\left(y+2\right);5\left(2-z\right)=3\left(y+z\right)\)và \(2x-3y+z=-4\). Tìm giá trị của \(B=x-y+z\)

Answer 1

\(\left\{{}\begin{matrix}2\left(x-3\right)=3\left(y+2\right)\\5\left(2-z\right)=3\left(y+2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2\left(x-3\right)}{6}=\dfrac{3\left(y+2\right)}{6}\\\dfrac{5\left(2-z\right)}{15}=\dfrac{3\left(y+2\right)}{15}\end{matrix}\right.\)

Hay \(\left\{{}\begin{matrix}\dfrac{x-3}{3}=\dfrac{y+2}{2}\\\dfrac{2-z}{3}=\dfrac{y+2}{15}\end{matrix}\right.\)

Tự làm được chứ?