Math 20 3 variable systems

I've tried this question so many times and I can't seem to get the right answer. Any takers??

Solve this system using substitution or elimination:

d+e-2f=-5

3d-e-4f=11

2d+4e+f=4

Here is what I did:

3(d+e-2f=-5)

__3d-e-4f=11__

3d+3f-6f=-15

-__3d-e-4f=11__

4e-2f=-26

2(d+e-2f=-5)

__4e-2f=-26__

2d+2e-4f= -10

-__2d+4e+f=4__

-2e-5f=-14

2(-2e-5f=-14)

__4e-2f=-26__

-4e-10f=-28

+__4e-2f=-26__

-12f=-54

f=-9/2

Then:

4e-2f=-26

4e-2(-9/2)=-26

4e + 9 = -26

4e = -35

e= -35/4

Then:

d+e-2f = -5

d+ (-35/4) -2(-9/2) = -5

d-35/4 + 9 = -5

d= -91/4

I have no idea where I went wrong. The answer is(33/4,-17/4,9/2)

So i got one variable right, but messed up on the others... ugh

Be careful with signs!!!!!

You obviously understand the process.

The line -12f = -54 implies f = +9/2.

Substitute +9/2 as you did with -9/2, and the remaining variables can be found.