1. Ta có: \(\frac{3+x}{5+y}=\frac{3}{5}\Leftrightarrow\hept{\begin{cases}3+x=3k\\5+y=5k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\left(k-1\right)\\y=5\left(k-1\right)\end{cases}}\)
\(\Rightarrow x+y=3\left(k-1\right)+5\left(k-1\right)=\left(3+5\right)\left(k-1\right)\)
\(\Rightarrow8\left(k-1\right)=16\)
\(\Leftrightarrow k-1=16\div8\)
\(\Leftrightarrow k-1=2\)
\(\Leftrightarrow k=2+1\)
\(\Leftrightarrow k=3\)
\(\Rightarrow\hept{\begin{cases}x=3.3-3=6\\y=5.3-5=10\end{cases}}\)
Vậy x = 6 và y = 10
Với \(\frac{3+x}{5+y}=\frac{3}{5}\Leftrightarrow x=3a;y=5a\left(1\right)\)
Ta có :
\(x+y=3a+5a\)
hay \(16=3a+5a\)
\(\Leftrightarrow16=8a\)
\(\Leftrightarrow a=2\left(2\right)\)
Thay ( 2 ) vào ( 1 ) . Ta có :
\(x=3.2;y=5.2\)
\(\Leftrightarrow x=6;y=10\)
Vậy x = 6; y=10
2. Ta có: \(\frac{x-7}{y-6}=\frac{7}{6}\Leftrightarrow\hept{\begin{cases}x-7=7k\\y-6=6k\end{cases}}\Leftrightarrow\hept{\begin{cases}x=7\left(k+1\right)\\y=6\left(k+1\right)\end{cases}}\)
\(\Rightarrow x-y=7\left(k+1\right)-6\left(k+1\right)=\left(7-6\right)\left(k+1\right)\)
\(\Rightarrow k+1=-4\)
\(\Leftrightarrow k=-4-1\)
\(\Leftrightarrow k=-5\)
\(\Rightarrow\hept{\begin{cases}x=7.\left(-5\right)+7=-28\\y=6.\left(-5\right)+6=-24\end{cases}}\)
Vậy x = -28, y = -24
