\(x+y=11=>y=11-x\left(1\right)\)
có: \(\dfrac{x}{10}+\dfrac{y}{15}=1\left(2\right)\)
thế(1) vào(2)=>\(\dfrac{x}{10}+\dfrac{11-x}{15}=1< =>\dfrac{3x+22-2x}{30}=1\)
\(=>x+22=30=>x=8\)(3)
thế (3) vào(1)=>y\(=11-8=3\)
\(\left\{{}\begin{matrix}x+y=11\\\dfrac{1}{10}x+\dfrac{1}{15}y=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{10}x+\dfrac{1}{10}y=\dfrac{11}{10}\left(1\right)\\\dfrac{1}{10}x+\dfrac{1}{15}y=1\left(2\right)\end{matrix}\right.\)
Lấy \(\left(1\right)-\left(2\right)\Rightarrow\dfrac{1}{30}x=\dfrac{1}{10}\Rightarrow x=3\Rightarrow y=11-3=8\)
