Question

College Algebra: I'm doing 3X3 Linear Systems and I'm having trouble solving this one. It's like everything I'm doing isn't working. The equation is:
2x+y-2z=-15
4x-2y+z=15
x+3y+2z=-5

Answer 1

notice how the last equation has x+3y+2z=-5 where the coefficient on the x is 1

this allows us to easily isolate x without leading to fractions

Let's solve x+3y+2z=-5 for x


x+3y+2z=-5

x+3y+2z-3y=-5-3y

x + 2z = -5 - 3y

x + 2z - 2z = -5 - 3y - 2z

x = -5 - 3y - 2z


making sense so far?

Answer 2

Alright, I understand that so far

Answer 3

now you take that and plug it into the other equations


plug x = -5 - 3y - 2z into the first equation


2x+y-2z=-15

2(-5 - 3y - 2z)+y-2z=-15 ... replace x with -5 - 3y - 2z

-10-6y-4z+y-2z = -15

-5y-6z-10 = -15

-5y-6z-10+10 = -15+10

-5y-6z = -5

Answer 4

so 2x+y-2z=-15 turns into -5y-6z = -5 after you plug in x = -5 - 3y - 2z and simplify

Answer 5

what happens when you do the same for 4x-2y+z=15 ?

Answer 6

Give me a second

Answer 7

Sure thing. Take your time.

Answer 8

So I plug x=-5-3y-2z into the second equation?

Answer 9

correct

Answer 10

replace all of x with -5 - 3y - 2z

Answer 11

I got -14y-7z=35

Answer 12

me too

Answer 13

so you now have this simplified 2x2 system

-5y-6z = -5
-14y-7z=35

do you see how to solve that?

Answer 14

The way I see it; I need to make one of the equation be y=something or z=something. Then plug that into one of the equations

Answer 15

I don't see how without getting a fraction

Answer 16

yes, essentially you are eliminating variables to try to get an equation of 1 variable so you can solve for that

Answer 17

which method do you want to use? elimination? or substitution?

Answer 18

So doing that, I'll get 3 answers? Honestly, I don't see how I'm going to get 3 solutions from doing this

Answer 19

you should get one solution per variable (x, y, z)

Answer 20

were you able to solve

-5y-6z = -5
-14y-7z=35

?

Answer 21

Yeah that's what I meant. So, what do I do next? I'm not sure how to solve this.

Answer 22

let's use elimination. That may be easier

Answer 23

which variable do you want to eliminate? y or z?

Answer 24

I'll say z

Answer 25

ok to eliminate z, we need something like -6z and +6z so they add to 0z and go away

but we have -6z and -7z, so they don't add to 0z like we want them to

Answer 26

but we can fix that.

we can multiply the -6z by 7 to get -42z

then we can multiply the -7z by -6 to get +42z

now they finally add to 0z

Answer 27

see how I did that?

Answer 28

Wait, yes I do.