Question

4x+3y+2z=43
2x+4y+3z=45
3x+2y+4z=47

Answer 1

i will be bak in 5 min

Answer 2

k..im back

Answer 3

now i need help

Answer 4

wat long replys

Answer 5

x = 13/3 , y=13/3 , z = 19/3 xDD

Answer 6

Probably Saifi is stuck he doesn't know how to do it :P

Answer 7

howd u get the answers wen u dunt even know how to do them @ H.mini

Answer 8

lol

Answer 9

LOL, I used my brain and i came up with the answers :P

Answer 10

Hey can you do it like the normal one with only 2 equations >

Answer 11

well at the back of the book it says (7, 4, 9) answers

Answer 12

http://www.wolframalpha.com/input/?i=4x%2B3y%2B2z%3D43+%2C2x%2B4y%2B3z%3D45%2C+3x%2B2y%2B4z%3D47

Answer 13

well..i dunno whch is wrong...

Answer 14

4x+3y+2z=43 [1]
2x+4y+3z=45 [2]
3x+2y+4z=47 [3]

To eliminate the z-variables from equations [2] and [3]...

Make a linear combination of
3(4x+3y+2z)=3*43 [1]'
-2(2x+4y+3z)=-2*45 [2]'
================
[2]''

Now make a linear combo of these
-2(4x+3y+2z)=-2*43 [1]''
3x+2y+4z=47 [3]
==============
[3]'

Then solve the 2x2 system involving [2]'' and [3]' for x and y then back substitute into one of the original euqations for z.

I checked the solutions on wolfram as did mimi, so i won't post them.

Answer 15

so is wolfram correct?

Answer 16

Well, use the answer you have and substitute it into the equation to see if its correct

Answer 17

I just checked the wolfram link, and those are the same answers i got; it is not likely we might have made the same typo. wolfram is probably correct; is it possible that you have a typo?

Answer 18

i didnt get these answers..i got them from the answers at the back of my book and they read (7, 4, 9)

Answer 19

Substitute it back in and see, there might be 2 answers to it

Answer 20

take mimi's advice it should be easy to check those integer answers in the equations to see if they satisfy

Answer 21

oh wait..i wrote the question wrong..the first equation is 4x+3y+2z=34

Answer 22

LOL, wonder why

Answer 23

the ol' typo on the question routine, hmmm...

Answer 24

so srry...

Answer 25

Let me try it, from what mandolino did xD

Answer 26

yep..wolfram is correct too wen i tried it..it gave (1 4 9)

Answer 27

Woah~ this is hard, im getting confused xD
Saifi got scared that he ran away

Answer 28

lol..maybe he did..anyways..i got like 30 problems like this...

Answer 29

Sorry I can't help.

Answer 30

dang it...
I HATE MATH!!!!

Answer 31

Me too *high five*

Answer 32

hero...i made a typo in the first equation its is: 4x+3y+2z=34

Answer 33

Hmmm. I'm talking to Lana right now, lol

Answer 34

i dunno who lana is...but i need a good stragety to get through these problems

Answer 35

wahhh..i need help!!!

Answer 36

Make a linear combination of
12x+9y+4z=102 [1]' x3
-4x-8y-6z=-90 [2]' x -2
================
8x+y =12 [2]''

Now make a linear combo of these
-8x-6y-4z=-86 [1]'' x -2
3x+2y+4z=47 [3]
==============
-5x-4y =-39 [3]'

Now solve the 2x2 system
8x+y =12 [2]''
-5x-4y =-39 [3]'

Mult [2]'' by 4 and make a linear combo
32x+4y=48 [2]''' x4
-5x-4y=-39 [3]'
==========
27x =9
x=9/27 = 1/3

I've got a mistake somewhere, but I can't find it. Maybe if I post it someone will find it. It's an arithmetic error. This is a very tedious problem.

Answer 37

There is no good strategy. You have to do these the hard way

Answer 38

can u esplain this way

Answer 39

It's a lot of steps

Answer 40

i really need help!!!!

Answer 41

Can you say that with a smile on your face?

Answer 42

I found the error; I went back to the bloody 43 instead of the 34 in the second linear combo to get [3]' I will redo from there now next post.

Answer 43

wat post? do i make another?

Answer 44

bloody math?

Answer 45

Now make a linear combo of these
-8x-6y-4z=-68 [1]'' x -2
3x+2y+4z=47 [3]
==============
-5x-4y =-21 [3]'

Now solve the 2x2 system

8x+y =12 [2]''
-5x-4y =-21 [3]'

Mult [2]'' by 4 and make a linear combo
32x+4y=48 [2]''' x4
-5x-4y=-21 [3]'
==========
27x =27
x=1

There must be another &^%#$% error! since x=13/3

Answer 46

NOPE..no errors..im jus soooo confused!!!

Answer 47

Holy &^$$ I am correct; I was using the old answer from wolfram!!!!!!

x=1, y=4, z=9

Answer 48

i still dunt understand ur method...could u kinda esplain it to me? like all the way till every answer?

Answer 49

Ok. a 3x3 is pretty daunting, so I used elimination to eliminate the same variable (z) from two of the equations [2] and [3]. That's the part where I mutliplied equation [1] and [2] so that the coefficients for z were additive inverses; I did the same thing with [1] and [3]. That's what I called the linear combination.
(i will continue on my next post)

Answer 50

help?

Answer 51

wat other post?

Answer 52

the one that i'm writing while you are reading what i just wrote

Answer 53

oh..

Answer 54

That gives us the 2x2 system involving equations [2]'' and [3]'. again i decided to use elimination method (but you could just as easily used substitution). in this case i decided to eliminate the y-variable by multiplying [2]'' by 4-- that made the coefficient for y in those two equations opposites so that when I added them together, the y was eliminated and I could solve for x.

Answer 55

my head hurts...

Answer 56

At this point, you can sub x=1 into any of the equations in the 2x2 to get y; i choose [2]''...

8(1)+y =12 [2]''
y=12-8=4

Now we have x=1 and y=4

Answer 57

To get z we have to sub those values for x and y into one of the given equations because it has a z variable in it. i choose equation [3]:

3(1)+2(4)+4z=47 [3]
3+8+4z=47
4z=47-11
4z=36
z=9
Finally we have x=1, y=4, z=9

Answer 58

i got the last part..trying to get the first 3 posts into my head..

Answer 59

This was an especially tedious 3x3 problem; I saw that from the beginning which is why I only wanted to set it up, and not solve it. The typo did not help. But these are often not this bad. it is better to learn on a less tedious problem.

Answer 60

I suggest that you make notes, or copy and paste; then try to redo the problem from the start when you are fresh.

Answer 61

ya..maybe i will do the rest tmmrw....

Answer 62

good idea

Answer 63

oh... thanks for the help hero

Answer 64

thanks a bunch though...

Answer 65

Me? What did I do?

Answer 66

ur welcome :})

Answer 67

just kidding around, lol

Answer 68

I would have helped but I was too busy trying to find Lana

Answer 69

who is this lana?

Answer 70

I can't tell you

Answer 71

yea who is lana when she's at home

Answer 72

oooohhhh!!!!