Question

Tìm x,y sao cho

\(\frac{x+4}{6}=\frac{3y-1}{8}=\frac{3y-x-5}{x}\)

Answer 1

Giải:

Ta có: \(\frac{x+4}{6}=\frac{3y-1}{8}=\frac{3y-x-5}{x}\)

\(\Rightarrow\frac{3y-1-x-4}{8-6}=\frac{3y-x-5}{x}\)

\(\Rightarrow\frac{3y-x-5}{2}=\frac{3y-x-5}{x}\)

+) \(3y-x-5=0\Rightarrow x+4=3y-1=0\)

+ \(x+4=0\Rightarrow x=-4\)

+ \(3y-1=0\Rightarrow3y=1\Rightarrow y=\frac{1}{3}\)

+) \(x=2\)

\(\Rightarrow x+4=3y-1\)

\(\Rightarrow2+4=3y-1\)

\(\Rightarrow3y=7+1\)

\(\Rightarrow3y=8\)

\(\Rightarrow y=\frac{8}{3}\)

Vậy cặp số \(\left(x,y\right)\) là: \(\left(-4;\frac{1}{3}\right);\left(2;\frac{8}{3}\right)\)