Question

\(\frac{3x+1}{18}+\frac{2y}{12}=\frac{2}{9}\)

và \(x-y=-1\)

Answer 1

Ta có \(\frac{3x+1}{18}+\frac{2y}{12}=\frac{2}{9}\)

\(\Rightarrow\frac{3x+1}{18}+\frac{y}{6}=\frac{2}{9}\Rightarrow\frac{3x+1}{18}+\frac{3y}{18}=\frac{2}{9}\)

\(\Rightarrow\frac{3x+3y+1}{18}=\frac{2}{9}\Rightarrow\frac{3.\left(x+y\right)+1}{18}=\frac{2}{9}\)

\(\Rightarrow9.\left[3.\left(x+y\right)+1\right]=36\Rightarrow3.\left(x+y\right)+1=4\)

\(\Rightarrow3.\left(x+y\right)=3\Rightarrow x+y=1\)

Mà \(x-y=1\Rightarrow x=y+1\)

Thay \(x=y+1\)vào \(x+y=1\)ta có

\(y+1+y=1\Rightarrow2y=0\Rightarrow y=0\)

Do đó \(x=1\)

Vậy x = 1 ; y = 0