Question

What is the solution to the system?
*PICTURES ATTACHED*

Answer 1

(-4, 0, 12)
(-3, 5, 6)
(0, -2, -12)
(2, 1, -3)

Answer 2

@yb1996 HELP PLEASE:(

Answer 3

First isolate z in the third equation. What do you get?

Answer 4

um i would isolate it by multiplying the first equation by -2... correct?

Answer 5

nope, just solve for z in the third equation

Answer 6

focus on just the third equation for now

Answer 7

ok one sec

Answer 8

idk how to do this:(

Answer 9

@jim_thompson5910

Answer 10

6x + z = -12

Answer 11

think of 6x + z as z + 6x

Answer 12

you must do something to both sides to move that 6x over

Answer 13

would i subtract the six on both sides or divide...something like that?

Answer 14

6x

Answer 15

subtract 6x from both sides

Answer 16

so now i have
z = -12 - 6x
correct?

Answer 17

yes

Answer 18

now that you know z = -12 - 6x, you can plug this into the other copies of 'z' in the other equations

Answer 19

for example, in equation 1

x + 3y - z = 6

x + 3y - ( z ) = 6

x + 3y - ( -12 - 6x ) = 6 ... replace every copy of z with -12 - 6x

x + 3y + 12 + 6x = 6

7x + 3y + 12 = 6

7x + 3y = 6 - 12

7x + 3y = -6

Hopefully you understand what is going on with each step

Answer 20

i kinda understand but I am not getting any closer to finding my answer...

Answer 21

we're getting there (eventually)

Answer 22

what do you get if you plug z = -12 - 6x into equation 2 ?

Answer 23

i give up thanks for the help though.

Answer 24

come on, give it a shot at least

Answer 25

4x - 2y + 2z = -10

replace z with -12 - 6x and simplify