Solve System of equation using elimination? \begin{align}
I:&& ~~ x+\frac12y &= 6
\\[.5em]
II:&& ~~ \frac32x + \frac{3}{2}y &= {17 \over 2}
\end{align}
when $x$ was multiplied by $(-3/2)$ in first equation the $x$ will be canceled and the resulting $y = -2/3$ and $x = 19/3$.
But when fractions were simplified first the resulting equation is
$$-8x-4y= -48$$
$$9x+4y = 51$$
here $y$ get canceled and results will be $x=3 , y=6$.
why these two attempts give two different results..I only need answer in elimination technique. 

This can solved in an entirely algorithmic way, calculating the RREF of the augmented matrix. First we multiply the second equation by $2$. Then:
\begin{align}
\begin{bmatrix}1&\frac 12& 6 \\ 3&3&17
\end{bmatrix}&\rightsquigarrow
\begin{bmatrix}1&\frac 12&\phantom{-}6 \\ 0&\frac32 &-1
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&\frac 12&\phantom{-}6 \\ 0& 1 &-\frac23
\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1& 0 &\color{red}{\frac{19}3} \\ 0& 1 &\color{red}{-\frac{2}3}
\end{bmatrix}
\end{align}
