Ta có \(\left(\frac{x}{y}\right)^2=\frac{16}{9}=\left(\pm\frac{4}{3}\right)^2\)
\(\frac{x}{y}\)dương nên \(\frac{x}{y}=\frac{4}{3}\Rightarrow x=\frac{4y}{3}\)
Thay \(x=\frac{4y}{3}\)vào \(x^2+y^2=100\)ta được
\(\left(\frac{4y}{3}\right)^2+y^2=100\)
\(\frac{16}{9}.y^2+y^2=100\)
\(y^2.\left(\frac{16}{9}+1\right)=100\)
\(y^2.\frac{25}{9}=100\)
\(y^2=100:\frac{25}{9}=36\)
\(y=6\)( vì y dương )