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. 

Multiplying the second equation by $\frac{2}{3}$ we get
$$x+\frac{1}{2}y=6,$$
and
$$x+y=\frac{17}{3}.$$
Now multiplying the first equation by $-1$ and adding to the second
$$\frac{1}{2}y=-6+\frac{17}{3}$$
Can you finish?
