Question

Solve the following system of equations:

2x + 3y – z = 1
3x + y + 2z = 12
x + 2y – 3z = –5 (5 points)


(3, 1, 2)
(–3, 1, 2)
(3, –1, 2)
(3, 1, –2)

Answer 1

@wio I got A am I right?

Answer 2

heheh

Answer 3

makes one wonder how A came about

Answer 4

Well I just plugged in the points and they worked for A so I want to see what others got to make sure I'm right

Answer 5

if A makes them all true, then A it is :)

Answer 6

hmmm I guess that's one way to go about it

Answer 7

hint: try to substitute the values in A into your system

Answer 8

Did you get A as well?

Answer 9

just bear in mind, the choices are there as guidelines, not as a menu to pick from

chicken or beef or turkey or whatever else

Answer 10

rref{{2,3,-1,1},{3,1,2,12},{1,2,-3,-5}}

im not getting A

and yes, trial and error is a valid method.

Answer 11

the points can be plugged in for x, y, and z respectively? and are you sure A's points don't work?

Answer 12

show your work, plug them in and ill show you your error

Answer 13

2(3) + 3(-1) – (2) = 1
6-3-2=1
3-2=1
1=1

3x + y + 2z = 12

x + 2y – 3z = –5



(3, –1, 2)
(3, 1, –2)

Answer 14

so A is not y=-1

Answer 15

right, so it's not A or B. So is it C or D?

Answer 16

i agree its C or D

Answer 17

how do i figure out if it's C or D?

Answer 18

well, the same way your approach is working, plug in the values and chk

Answer 19

plug in A, it fails
plug in B it fails
plug in C ... ???

2(3) + 3(-1) – (2) = 1

3(3) + (-1) + 2(2) = 12

(3) + 2(-1) – 3(2) = –5




6-3-2 = 1

9-1+4 = 12

3-2-6 = –5


are these true?

Answer 20

yes, the values for C are true.

Answer 21

umm, no. they have to equal what they equal .... their equality is already defined by the equations that exist in the system.

then since C fits, C must be the solution.

Answer 22

yeah sorry i made a mistake haha thank you

Answer 23

good luck :)