Đáp án: $a$) `(x;y)=(2;-1)`.
$b$) `(x;y)=(1;-5)`.
$c$) `(x;y)=(16/7;-19/7)`.
Giải thích các bước giải:
$a$) $\left\{\begin{matrix}4x+5y=3& \\x-3y=5&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}4x+5y=3& \\4x-12y=20&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}4x+5y-4x+12y=3-20=-17& \\4x-12y=20&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}17y=-17& \\x-3y=5&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}y=-1& \\x-3y=5&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}y=-1& \\x=2&\end{matrix}\right.$
Vậy `(x;y)=(2;-1)`.
$b$) $\left\{\begin{matrix}2x+y=-3& \\2x-3y=17&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}2x+y-2x+3y=-3-17=-20& \\2x-3y=17&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}4y=-20& \\2x-3y=17&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}y=-5& \\2x-3y=17&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}y=-5& \\x=1&\end{matrix}\right.$
Vậy `(x;y)=(1;-5)`.
$c$) $\left\{\begin{matrix}5x-5y=25& \\3x+4y=-4&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}5.(x-y)=25& \\3x+4y=-4&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}x-y=5& \\3x+4y=-4&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}3x-3y=15& \\3x+4y=-4&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}3x-3y-3x-4y=15-(-4)=19& \\3x+4y=-4&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}-7y=19& \\3x+4y=-4&\end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}y=\dfrac{-19}{7}& \\x=\dfrac{16}{7}&\end{matrix}\right.$
Vậy `(x;y)=(16/7;-19/7)`.
