Question

Solve the following system of equations.

x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1

a. (–3, –1, –5)
b. (–3, 5, –2)
c. (2, –4, –3)
d. (4, 4, –1)

Answer 1

do u know how to use cramer?

Answer 2

use substitution, or elimaination method and solve for 2 of the variables.

Answer 3

wasn't he that guy on seinfeld...?

Answer 4

Me have no idea what seinfeld is O.o

Answer 5

@bombshellbri do u know that or not? if u don't i can explain

Answer 6

Or try adding the equations together in a way to make one variable disappear. Do this twice and then you have a 2x2 system.

Answer 7

i dont not know

Answer 8

x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1

Add E2 + E3 to E1
5x + 3z = -30

Answer 9

@angela210793 what is cramer?

Answer 10

hmmm...well to use Cramer u must know matrices...do u know that? I'm insisting on Cramer cause it's very easy

Answer 11

im sorry i dont know it either

Answer 12

hmmm....well I guess @telliott99 can help u with his method...cause i make very masitakes in this method :/

Answer 13

mistakes*

Answer 14

alrite and thats okay

Answer 15

Yes. The trouble with Cramer's rule for 3 x 3 is there is a lot of arithmetic.

Answer 16

So we have one equation missing y. We need another one.

Answer 17

x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1
How about multiplying E2 by 3 and adding it to E1. I get
7x + 8z = -61
Before we had
5x + 3z = -30

Multiply the first one by -3 and the second one by 8
-21x -24z = 183
40x + 24z = -120

Add to get
19x = 63

That's pretty ugly. Maybe I made a mistake. Anyway, you see how to do that method.

Maybe I'll try getting rid of z instead

x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1

Add twice E3 to both E1 and E2
5x - y = 14
6x -5y = -17

So y = 5x - 14
6x - 5(5x -14) = -17
-19x + 70 = -17
x = -87/-19 = 87/19
still ugly

Answer 18

And one of those must have a mistake!

Answer 19

Sorry I didn't solve your problem. 3 x 3 are hard enough, and harder with larger numbers and fractions, etc. But this and Cramer's rule are the two usual methods.
http://en.wikipedia.org/wiki/Cramer's_rule

Answer 20

Oh, I forgot.

Answer 21

Since you have possible answers, you can always plug them into the equations.

Answer 22

x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1

a. (–3, –1, –5)
b. (–3, 5, –2)
c. (2, –4, –3)
d. (4, 4, –1)

Answer 23

Start with (d), they usually put the ones that work for most at the top.

Answer 24

but he said he doesn't know matrices...that's why i quit
and he can't do that Telliot....wht if he has 100 equations? is he really gonna try all the options O.o

Answer 25

(d) does not work for E1

Answer 26

im a girl* and does it matter which equation you use?

Answer 27

No for 100 equations you get a computer (and the computer uses my method).

Answer 28

We have eliminated (d) because it does not solve E1. Do you see that?

Answer 29

sorry
and that's not the right way to do it -_- she can't tell her teacher that she tried all the options to see which one satisifed all the equations :S

Answer 30

sorry E3

Answer 31

You wanna do Cramer's?

Answer 32

If the method of solution is not specified, checking the answers seems perfectly valid to me.

Answer 33

Particularly when they have not been taught methods of solution, apparently.

Answer 34

i dont know what cramer's is. im only in algebra 2 ....

Answer 35

Well, it's not too hard, if you want to try we can do it

Answer 36

Actually, it will be a lot of work for a 3x3

Answer 37

its an online class so we can try all the options to see which one satisifed all the equation

Answer 38

Bingo

Answer 39

I'm not telling to use cramer..there must be another way to solve this
but whtever...do wht u want :P

Answer 40

"must be"---there is not

Answer 41

(b) does not solve E3

Answer 42

Can you check the other two @bombshellbri

Answer 43

hey telliot99 can you help me?

Answer 44

using B?

Answer 45

If I did it right, I found that
(b) does not solve Eqn 3 and neither does (d)
That leaves (a) and c as possible solutions.

Answer 46

I think you can plug in the numbers to check?

Answer 47

okay, ill try them

Answer 48

Did you get an answer. I have one but I'm hoping you can tell me.

Answer 49

(2x – y + 2z) - (x + 3y + 2z) = -15 + 16
x - 4y = 1
x = 1+4y
(2x – 2y – z) - (2x – y + 2z) = 1 + 15
-y - 3z = 16
2(1+4y) - 2y - z = 2 + 6y - z = 1
6y - z = -1
18y - 3z = -3
-19y = 19, y = -1
x = -3
z = -5
answer: (A)

Answer 50

No, sorry, we are no longer adding equations together
x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1
we are just trying this as a solution
a. (–3, –1, –5)

I get -3 + 3(-1) + 2(-5) = -16 check
2(-3) - (-1) + 2(-5) = -15 check
2(-3) -2(-1) -(-5) = 1 check

Answer 51

Oh, I see you added them together correctly!

Answer 52

Great job!

Answer 53

thank you very much :D

Answer 54

yw