Question

\(\left\{{}\begin{matrix}2x-3y+1=0\\\dfrac{x+1}{2}=\dfrac{4y-1}{3}\end{matrix}\right.\)

Answer 1

\(\left\{{}\begin{matrix}2x-3y+1=0\\\dfrac{x+1}{2}=\dfrac{4y-1}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3y=-1\\3\left(x+1\right)=2\cdot\left(4y-1\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3y=-1\\3x+3-8y+2=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-3y=-1\\3x-8y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}6x-9y=-3\\6x-16y=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=7\\2x-3y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=1\\2x=3y-1=3\cdot1-1=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=1\\x=1\end{matrix}\right.\)