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. 
 A: 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?
A: Multiply the first equation by $3$ and the second by $2$. Now using the elimination method, I have: $-\frac{3}{2}y=1$ so $y=-\frac{2}{3}$. Substituing for $x$, I obtain: $x=6+\frac{1}{3}=\frac{19}{3}$.
A: Multiplying $I$ by $8$  yields $$-8x-4y=-48$$ and multiplying $II$ by $\dfrac83$ yields $$\color{red}4x+4y=\color{red}{\dfrac{68}3}.$$ 
Adding these together yields $$-4x=-48+{\dfrac{68}3}$$
or $$x=12-\dfrac{17}3=\dfrac{19}3.$$
A: 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}
