Question

\(\frac{x-3}{8}=\frac{y}{30}=\frac{z+1}{27}\); \(3x-5z+2y=23\)

Answer 1

\(\frac{x-3}{8}=\frac{y}{30}=\frac{z+1}{27}=\frac{3x-9}{24}=\frac{2y}{60}=\frac{5z+5}{135}=\frac{3x-9-\left(5z+5\right)+2y}{24-135+60}=\frac{3x-5z+2y-14}{-51}\)

\(=\frac{23-14}{-51}=\frac{9}{-51}=\frac{3}{-17}\)

\(\Rightarrow\hept{\begin{cases}x-3=\frac{8.3}{-17}=-\frac{24}{17}\Rightarrow x=\frac{-24}{17}+3=\frac{27}{17}\\y=\frac{30.3}{-17}=-\frac{90}{17}\\z+1=\frac{27.3}{-17}=-\frac{81}{17}\Rightarrow z=-\frac{81}{17}-1=-\frac{98}{17}\end{cases}}\)