\(2x=3y\Rightarrow x=3k;y=2k\left(k\in N\right)\Rightarrow3x-2y=9k-6k=3k=1,5\Rightarrow k=0,5\Rightarrow x=1,5;y=1\)
Từ \(2x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{2}\)
\(\Rightarrow\frac{x}{3}=\frac{y}{2}=\frac{3x}{9}=\frac{2y}{4}=\frac{3x-2y}{9-4}=\frac{2,5}{5}=0,5\)
\(\Rightarrow x=0,5.3=1,5\)
\(y=0,5.2=1\)
Vậy \(x=1,5\)và \(y=1\)
T a có \(2x=3y\Leftrightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{3x}{3.3}=\frac{2y}{2.2}\)
\(\Rightarrow\frac{3x}{9}=\frac{2y}{4}\)
áp dụng tính chất dãy tỉ số bằng nhau :
\(\frac{3x}{9}=\frac{2y}{4}=\frac{3x-2y}{9-4}=\frac{2.5}{5}=0.5\)
\(\Rightarrow\)\(3x=9.0,5=4,5\)\(\Rightarrow x=\frac{4,5}{3}=1,5\)
\(\Rightarrow\)\(2y=4.0,5=2\)\(\Rightarrow y=\frac{2}{2}=1\)
Vậu ta có (x;y) = (1,5 ; 1 )
