\(\left\{{}\begin{matrix}-2x+ay=4\\3x+y=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5-3x\\-2x+a\left(5-3x\right)=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5-3x\\-2x+5a-3xa=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-3x+5\\x\left(-3a-2\right)=4-5a\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-3x+5\\x\left(3a+2\right)=5a-4\end{matrix}\right.\left(1\right)\)
TH1: \(a=-\dfrac{2}{3}\)
(1) sẽ tương đương với \(\left\{{}\begin{matrix}y=-3x+5\\x\cdot0=5\cdot\dfrac{-2}{3}-4=-\dfrac{10}{3}-\dfrac{12}{3}=-\dfrac{22}{3}\left(vôlý\right)\end{matrix}\right.\)
=>Loại
TH2: a<>-2/3
(1): \(\left\{{}\begin{matrix}y=-3x+5\\x\left(3a+2\right)=5a-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5a-4}{3a+2}\\y=-3x+5=\dfrac{-3\cdot\left(5a-4\right)}{3a+2}+5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\dfrac{5a-4}{3a+2}\\y=\dfrac{-15a+12+15a+10}{3a+2}=\dfrac{22}{3a+2}\end{matrix}\right.\)
x>0 và y>0
=>\(\left\{{}\begin{matrix}\dfrac{5a-4}{3a+2}>0\\\dfrac{22}{3a+2}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3a+2>0\\\dfrac{5a-4}{3a+2}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a>-\dfrac{2}{3}\\\dfrac{5a-4}{3a+2}>0\end{matrix}\right.\)
\(\dfrac{5a-4}{3a+2}>0\)
TH1: \(\left\{{}\begin{matrix}5a-4>0\\3a+2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a>\dfrac{4}{5}\\a>-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow a>\dfrac{4}{5}\)
mà a>-2/3
nên \(a>\dfrac{4}{5}\)
TH2: \(\left\{{}\begin{matrix}5a-4< 0\\3a+2< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a< \dfrac{4}{5}\\a< -\dfrac{2}{3}\end{matrix}\right.\)
=>\(a< -\dfrac{2}{3}\)
mà a>-2/3
nên \(a\in\varnothing\)
Vậy: \(a>\dfrac{4}{5}\)
mà a là số nguyên nhỏ nhất
nên a=1
