Question

Still not getting this.
help!
I tried using the substitution method, it's not making sense. The problem says:
What is the solution of the system of equations?

3x + 2y + z = 7
5x + 5y + 4z = 3
3x + 2y + 3z =1

I get z = -3 (if that's right) then I'm lost

Answer 1

1st you need to take the 1st and 2nd equation.... choose to cancel out a variable. I will choose z.

(-4)3x + 2y + z = 7
5x + 5y + 4z = 3

-12x - 8y - 4z = -28
5x + 5y + 4z = 3
-----------------
-7x -3y = -25 (One Equation to come back and use)

Answer 2

ok

Answer 3

btw, z =-3, so that is correct.

Answer 4

ok that makes sense so far

Answer 5

I originally cancelled out y first

Answer 6

ok I'm going to do this on paper first okay?

Answer 7

ok

Answer 8

Okay good...

Answer 9

@tms622 I will do it your way. You were fine in your logic.

Answer 10

I got lost after I figured out z. Y and X confused me or something.

Answer 11

(5)3x+2y+z=7
(-2)5x+5y+4z=3
---------------
15x+10y+5z=35
-10x-10y-8z=-6
----------------
5x-3z=29 (One Equation)


(2)5x+5y+4z=3
(-5)3x+2y+3z=1
----------------
10x+10y+8z=6
-15x-10y-15z=-5
----------------
-5x-7z=1 (Another Equation)


5x-3z=29
-5x-7z=1
----------
-10z=30
z = -3

Answer 12

Now.... since we have 'z' we must go back to an equation that has only TWO variables. So either...

5x-3z=29

OR

-5x-7z=1

Answer 13

ok

Answer 14

Lets choose the all positive one. Then substitute in z.

5x-3(-3)=29
5x + 9 = 29
5x = 20
x = 4

Answer 15

Now we have TWO variables solved for...
z=-3
x= 4

Now we need an equation with 3 variables in it.. then substitute those two in there.

Answer 16

ohhh ok then solve for y

Answer 17

So we can choose out of.... the originals :)

3x + 2y + z = 7
5x + 5y + 4z = 3
3x + 2y + 3z =1

Answer 18

3(4)+2y-3=7

12 + 2y - 3 = 7
9+2y = 7
2y = -2
y = -1

Answer 19

yay Thats what I just got too. Thanks for all your help. That makes alot more sense now.

Answer 20

:)