Question

\(\hept{\begin{cases}\left(2x-3\right)\cdot\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\cdot\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\) giải hệ phương trình

Answer 1

\(\hept{\begin{cases}\left(2x-3\right)\left(2y+4\right)=4x\left(y-3\right)+54\\\left(x+1\right)\left(3y-3\right)=3y\left(x+1\right)-12\end{cases}}\)

\(\hept{\begin{cases}4xy+8x-6y-12=4xy-12x+54\\3xy-3x+3y-3=3xy+3y-12\end{cases}}\)

\(\hept{\begin{cases}4xy-4xy+8x+12x-6y-12-54=0\\3xy-3xy-3x+3y-3y-3+12=0\end{cases}}\)

\(\hept{\begin{cases}20x-6y-66=0\\-3x+9=0\end{cases}}\)

\(\hept{\begin{cases}2\left(10x-3y\right)=66\\-3\left(x-3\right)=0\end{cases}}\)

\(\hept{\begin{cases}10x-3y=33\\x-3=0\end{cases}}\)

\(\hept{\begin{cases}10x-3y=33\\x=3\end{cases}}\)