a: \(\left\{{}\begin{matrix}2x-3y=5\\4x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=10\\4x+y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-7y=7\\4x+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\4x=3+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Thay x=1 và y=-1 vào hệ còn lại, ta được:
\(\left\{{}\begin{matrix}2\cdot1-3\cdot\left(-1\right)=5\\12\cdot1+3\cdot\left(-1\right)=a\end{matrix}\right.\Leftrightarrow a=12-3=9\)
b: \(\left\{{}\begin{matrix}x-y=2\\3x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x=3\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{3}{4}-2=-\dfrac{5}{4}\end{matrix}\right.\)
Thay x=3/4 và y=-5/4 vào hệ còn lại, ta được:
\(\left\{{}\begin{matrix}2\cdot a\cdot\dfrac{3}{4}-2\cdot\dfrac{-5}{4}=1\\\dfrac{3}{4}-\dfrac{5}{4}a=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}a=1+2\cdot\dfrac{-5}{4}=1-\dfrac{5}{2}=-\dfrac{3}{2}\\\dfrac{5}{4}a=\dfrac{3}{4}-2=-\dfrac{5}{4}\end{matrix}\right.\)
=>a=-1
