# Thread: Math 20 3 variable systems

1. ## 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

2. ## 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.

3. oh my gosh.
how did i miss that??
thanks

4. Originally Posted by t-dot
Solve this system using substitution or elimination:
d+e-2f=-5 [1]
3d-e-4f=11 [2]
2d+4e+f=4 [3]
You seem to go to a lot of unnecessary work.

Can be much simpler; like:

Add [1] + [2] to get 4d - 6f = 6 ; 2d - 3f = 3 [4]

Multiply [2] by 4 : 12d - 4e - 16f = 44
Add to [3] to get 14d - 15f = 48 [5]

Multiply [4] by -7 : -14d + 21f = -21
Add to [5] to get 6f = 27 ; f = 9/2

Now get d and e.