Solve System of equation using elimination?

Question

\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.

Answer 1

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?

Answer 2

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: $x=6+\frac{1}{3}=\frac{19}{3}$.

Answer 3

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 $$x=12-\dfrac{17}3=\dfrac{19}3.$$

Answer 4

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}