Question

Tìm x,y,z biết: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và 2x+3y-z=32

Answer 1

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{2x+3y-z-2-6+3}{4+9-4}=\frac{27}{9}=3\)

\(\Rightarrow x=\frac{3\cdot4+2}{2}=7\)

\(\Rightarrow y=\frac{3\cdot9+6}{3}=11\)

\(\Rightarrow z=3\cdot4+3=15\)

Answer 2

       \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\Rightarrow\frac{2x-2}{4}=\frac{3x-6}{9}=\frac{z-3}{4}=\frac{2x+3y-z-2-6+3}{4+9-4}\)

= \(\frac{27}{9}\)

=  \(3\)

\(\Rightarrow\)\(x=\frac{3.4+2}{2}=7\)

\(\Rightarrow\)\(y=\frac{3.9+6}{3}=11\)

\(\Rightarrow\)\(z=3.4+3=15\)