\(2\left(x-3\right)=3\left(y+2\right)\Rightarrow\frac{x-3}{3}=\frac{y+2}{2}\Rightarrow\frac{x-3}{15}=\frac{y+2}{10}\)
\(5\left(2-z\right)=3\left(y+2\right)\Rightarrow\frac{2-z}{3}=\frac{y+2}{5}\Rightarrow\frac{2-z}{6}=\frac{y+2}{10}\)
\(\Rightarrow\frac{x-3}{15}=\frac{y+2}{10}=\frac{2-z}{6}=\frac{2x-6}{30}=\frac{3y+6}{30}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x-3}{15}=\frac{y+2}{10}=\frac{2-z}{6}=\frac{2x-6}{30}=\frac{3y+6}{30}=\frac{\left(2x-6\right)-\left(3y+6\right)-\left(2-z\right)}{30-30-6}=\frac{\left(2x-3y+z\right)+\left(-6-6-2\right)}{-6}=\frac{-4-2}{-6}=1\)
=> x - 3 = 15.1 = 15 => x = 18
y + 2 = 10.1 = 10 => y = 8
2 - z = 6.1 = 6 => z = -4
→ 2(x-3) =3(y+2) = 5( 2-z)
→ x-3/15 =y+2/10=2-z/6
→ 2x-6/15=3y+6/10=2-z/6=-18/-6=3
→ x=48; y= 28 ; z= -16
