For problem number 1, simply set up a system and solve for x, y and z.

\(\displaystyle 3x-2y+2z=1\)

\(\displaystyle x+2y-z=5\)

\(\displaystyle 4x-3y+z+-3\)

Here, add the first two equations to get:

\(\displaystyle 4x+z=6\)

Now subtract the third equation to get:

\(\displaystyle 3y=9\)

\(\displaystyle y=3\)

Now, plug this in and solve for the others. I won't do everything for you.

For number 2, I suggest what dwsmith suggested.

For number 3 a., this is a simple system of equations. Start of with the first equation and simplify (the 2y/2 becomes just y, because the 2's will cancel out:

\(\displaystyle x+y=-0.25y\)

\(\displaystyle x=-1.25y\)

Now plug this back into the second equation:

\(\displaystyle -1.25y +y =7\)

\(\displaystyle -0.25y=7\)

\(\displaystyle y=-28\)

therefore :

\(\displaystyle x-28=7\)

\(\displaystyle x=35\)

\(\displaystyle y=28\)

Now as for part B, i believe that you cannot solve this system, because the equations given are essentially the same. The second equation is just the first one multiplied by -1.5

This link might help as well:

Systems of Linear Equations: Solving by Substitution