solve the following using augmented matrix's
4x-7y=-2
x+2y=7
(pls show work so i can understand thx)

- anonymous

solve the following using augmented matrix's
4x-7y=-2
x+2y=7
(pls show work so i can understand thx)

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- amistre64

[4 -7 | -2]
[1 2 | 7]

- amistre64

matrix is just the elimination process without the variables getting in the way

- amistre64

[4 -7 | -2]
[1 2 | 7] <- (*-4)
[ 4 -7 | - 2]
[-4 -8 | -28] <- (subtract R2 from R1)
[ 4 -7 | - 2]
[-4 -8 | -28]
-----------
[0 -15 | -30] <- (/-15) to get the 2nd number to be a 1
[0 1 | 2] <- new R2

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## More answers

- amistre64

[ 4 -7| - 2] <- (/4) to get teh 1st number to be 1
[0 1 | 2]
[1 -7/4| - 1/2]
[0 1 | 2 ] <-(*7/4) and subtract from R1

- amistre64

[1 -7/4| - 1/2]
[0 7/4 | 7/2 ]
--------------
[1 0 | 6/2] <- new R1, reduce 6/2 to 3
[1 0 | 3]
[0 1 | 2]

- amistre64

with any luck; x=3 and y = 2

- amistre64

4(3)-7(2)=-2
12 - 14 = -2 checks
(3)+2(2)=7
3 + 4 = 7 checks

- amistre64

it is the same process as elimination; since the variables dont matter anyways; we can just strip them out of the way and work with the numbers instead

- anonymous

dear god thank you tho i still dnt quite understand how to do the other 4 :(

- amistre64

post them lol

- anonymous

k thank you

- amistre64

instead of trying to hunt down the [,|, and ] buttons im gonna leave them out ot it ;)

- anonymous

alrighty thanks want me to post here or under ask a question?

- amistre64

youre call ;)

- anonymous

btw you are saying my graduation i have all the other parts of my precal credit recovery but this haha i wish i could thank you in person my teachers at school scratched their heads and said good luck thank you

- anonymous

ill post here

- amistre64

:)

- anonymous

11. 5x=3y-50
2y=1-3x
13. x+y+z=-2
2x-3y+z=-11
-x+2y-z=8

- amistre64

do you know how to solve a system of equations by elimination?

- anonymous

if i saw the problem i could tell you but i dnt remember alot of precal vocab

- amistre64

5 -3 = -50 ; Row1(R1) should be gotten to a 1 0 format
-3 2 = 1 ; Row2(R2) should be gotten to a 0 1 format here

- amistre64

5 -3 = -50 ; if we /5 we get a 1 x = x right?
-3 2 = 1
1 -3/5 = -10 ; *3 to eliminate the -3 in R2
-3 2 = 1
3 -9/5 = -30
-3 2 = 1 <- add them together
------------
0 1/5 = -29 <- *5 to turn that 1/5 unto a 1
0 1 = -145 <- this tells us y = -145 if i did it right lol

- anonymous

the ansers shld come out to be x= -97/19 and y= 155/19 in the back of the book

- amistre64

1 -3/5 = -10 ; new R1
0 1 = -145 ; new R2 <- (*3/5) to eliminate the -3/5 in R1
1 -3/5 = -10
0 3/5 = -87 <- add them toghter
-------------
1 0 = -97
but i could have typos somewhere :)

- amistre64

let me recheck without the explanations ok....

- amistre64

5x -3y=-50
3x+2y = 1 <- had a -3 there...bad me lol

- anonymous

its fine man

- amistre64

5 -3 = -50 (/5)
3 +2 = 1
1 -3/5 = -10 ; *-3
3 +2 = 1
-3 + 9/5 = +30
3 +10/5 = 1
---------------
0 19/5 = 31 (5/19)
0 1 = 151/19 ??

- amistre64

31
5
---
155 lol 155/19

- amistre64

1 -3/5 = -10
0 1 = 155/19 (3/5)
1 -3/5 = -10
0 3/5 = 93/19
---------------
1 0 = -190+93/19
190
93
----
97/19
1 0 = -97/19
0 1 = 155/19 right?

- amistre64

its easier on the paper lol

- amistre64

does it look any easier yet :)

- anonymous

starting to but i look at these problems and i just feel myself getting lost :(

- amistre64

1 1 1 = - 2
2 -3 1 = -11
-1 2 -1 = 8

- amistre64

when we look at this the 1st and 3rd are ready to be added togeter and eliminate that -1 right

- anonymous

oh
the first and third 1 tho

- anonymous

dnt we need the bottom right to be 1?

- anonymous

i guess we do that later?

- amistre64

1 1 1 = - 2
-1 2 -1 = 8
------------
0 3 0 = 6 (/3)
0 1 0 = 2 ..... y = 2

- amistre64

we use R1 to calibrate R2 and R3

- anonymous

but hte third row is z?

- anonymous

so dnst it also need it to be 0 0 1?

- amistre64

1 1 1 = - 2 (*-2)
2 -3 1 = -11
-2 -2 -2= 4
2 -3 1 = -11
--------------
0 -5 -1 = -7 <- new R3

- amistre64

we dont have to keep the rows in order; just as long as we pick an order and stick to it :) to keep things organized for us lol it just so happened that R3 made a perfect R2 :)

- amistre64

0 1 0 = 2 (*5)
0 -5 -1 = -7
0 5 0 = 10
0 -5 -1 = -7
-------------
0 0 -1 = 3 (*-1)
0 0 1= -3 z=-3

- amistre64

1 1 1 = - 2
0 1 0 = 2
0 0 1 = -3
1 2 -3 = -2
1 0 0 = -2 -2 3
1 0 0 = -1 x = -1 right?

- anonymous

wait when did R1 become 1 2 -3 ?

- amistre64

1 0 0 = -1
0 1 0 = 2
0 0 1 = -3
1 1 1 = -2 : mean x+y+z = -2
x +2 - 3 = -2
x = -2-2+3
x = -1

- anonymous

oh ok

- amistre64

whenever we know what a '1' in a column equals; we can substitute it to find aonther :)

- anonymous

see i didnt think you could so yet another reason why i couldnt self teach myself this
hey do you want to keep going or food break?

- amistre64

lol..... how many more you got?
just remember that we stripped the variables and that a 1 = the variable
0 1 0 = 2 means!! 1y=2, y=2

- anonymous

two more of this type and 1 more then IM DONE HURRAY
and ya that makes sense just couldnt wrape my head around the concept im working 6 hours math after school with about four to three hours of sleep a night its catching up too me :(

- amistre64

x y z = #
---------
1 0 0 = -1
0 1 0 = 2
0 0 1 = -3

- anonymous

i dont even get the option of failing this cause my navy nuke recruiter wld kill me :(

- amistre64

lol.... i dont know much about matrixes, but i do know this :)

- amistre64

give me another

- amistre64

linear algebra has lots of matrix stuff in it :)

- amistre64

and maybe discrete mathematics

- amistre64

post a new question box so we dont bog this post down ...long posts tend to slow my system ;)

- anonymous

k ill throw up the last two of these and grab a snack and brb here they are

- anonymous

ok then ill do that

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