Ido not understand this at all, and they get worse. equations are hard for me, fractions are hard for me and don't start me with integers, but i digress. my question is this, how do you solve these types of problems? i need some one with a finite math background. i mean really knows this stuff. can i get with my class. it's 4 weeks long and only 2 assignment due a week.
2x-6z=1
x+3y+2z=3
3y+8z=4

These problems take a little longer as you have to solve for more than one variable but her is the idea:
I want to get only one variable so I am going to find what other variables are in terms of that variable(z is in all three so it would be easiest to do this problem by finding the others in terms of it)
if 2x-6z=1 add 6z to both sides to give you 2x=1+6z
divide by 2 to get x=1/2+3z

we do the same with 3y+8z=4
subtract 8z from both sides to get 3y=4-8z
then divide by 3 to get y=4/3-8/3z 

Now I will plug them both into my middle equasion (x+3y+2z=3)
to get (1/2+3z)+3(4/3-8/3z)+2z=3 
the threes cancel out and because all we have is addition and subtraction you can now get rid of the parenthesis 1/2+3z+4-8z+2z=3 
combine like terms to get 9/2-3z=3 (i wrote the 4 1/2 as 9/2)
subtract 9/2 to get -3z=-3/2
multiply by a negative to get 3z=3/2
Divide by 3 and you get z=1/2

Now you have to find the other two variables to do this I would just plug z into the other equations a solve

2x-6z=1
2x-6(1/2)=1 plug in z to get 2x-3=1
add the three to both sides to get 2x=4
divide by two to get x=2

Finally the last equation:
3y+8z=4
plug in z to get 3y+8(1/2)=4 or 3y+4=4
subtract the 4 from both sides to get 3y=0
divide by 3 and you get y=0

So overall we have z=1/2, x=2, and y=0

To check that we did nothing wrong I would recommend plugging it back into our equation (use one that includes all three variables)
x+3y+2z=3
(2)+3(0)+2(1/2)=3
simplify 2+0+1=3  So your values should be correct.

Hope this was helpful! :)

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