Originally Posted by
Quick hmm...
first let's distribute out the equations to get:
$\displaystyle 3(x-y)=2(y-x) + 8\quad\rightarrow\quad 3x-3y=2y-2x + 8$
$\displaystyle -(2x - y) = 3(y + 3x - 1)\quad\rightarrow\quad -2x+y=3y+9x-3$
Alright, now let's solve for x (or y, it's your choice) for the first problem:
$\displaystyle 3x-3y=2y-2x + 8\quad\rightarrow\quad 5x=5y+8\quad\rightarrow\quad x=\frac{5y+8}{5}$
now take the second equation:
$\displaystyle -2x+y=3y+9x-3$
and also solve for x:
$\displaystyle -2x+y=3y+9x-3\quad\rightarrow\quad -11x=2y-3\quad\rightarrow\quad x=\frac{2y-3}{-11}$
and now we can start solving stuff. First things first, you have to write down that:
$\displaystyle x=x$
So now we substitute:
$\displaystyle \overbrace{\frac{5y+8}{5}}^{\text{from the first equation}}=\underbrace{\frac{2y-3}{-11}}_{\text{from the second equation}}$
so now multiply both sides by -11 and 5 to get:
$\displaystyle -11(5y+8)=5(2y-3)$
now use the distributive property:
$\displaystyle -55y-88=10y-15$
now move things around:
$\displaystyle -65y=73$
then divide:
$\displaystyle y=\frac{-73}{65}$
Now you can solve for x.
On a side note, $\displaystyle \frac{-73}{65}$ is not a nice number, you should double check my work and see if my arithmetic is right.