Solve for each of the variables:

3(x+3)/5 +y/2 + 2z = -1

x/3 + 5y-2/4 + z= 0

x - 2(y+1)/3 - 3z =-2

Clear fractions to get things easier to work with:

6(x+3)+5y+20z = -10
4x+60y-6+12z = 0
3x-2(y+1)-9z = -6

Clear parentheses and collect variables on the left to get

6x+5y+20z = -28
4x+60y+12z = 6
3x-2y-9z = -4

Now just solve by your favorite method to get some nasty fractions.

If, however, you are less careless with your parentheses, and the second equation is

x/3 + (5y-2)/4 + z= 0

then you get

(x,y,z) = (-3, 2, -1)

There are no parantheses in the second equation its just:

x/3 + 5y-2/4 + z =0

yes, but if it says

5y-2
-------
4

That means that the quantity (5y-2) is divided by 4.

If not, then go for the messy fractions.

It's just that I cannot believe that they'd provide an equation with 2/4 as one of the terms.