Question

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

Answer 1

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

Answer 2

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

Answer 3

[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

Answer 4

[ 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

Answer 5

[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]

Answer 6

with any luck; x=3 and y = 2

Answer 7

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


(3)+2(2)=7
3 + 4 = 7 checks

Answer 8

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

Answer 9

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

Answer 10

post them lol

Answer 11

k thank you

Answer 12

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

Answer 13

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

Answer 14

youre call ;)

Answer 15

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

Answer 16

ill post here

Answer 17

:)

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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 :)

Answer 25

let me recheck without the explanations ok....

Answer 26

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

Answer 27

its fine man

Answer 28

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 ??

Answer 29

31
5
---
155 lol 155/19

Answer 30

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?

Answer 31

its easier on the paper lol

Answer 32

does it look any easier yet :)

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

oh
the first and third 1 tho

Answer 37

dnt we need the bottom right to be 1?

Answer 38

i guess we do that later?

Answer 39

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

0 1 0 = 2 ..... y = 2

Answer 40

we use R1 to calibrate R2 and R3

Answer 41

but hte third row is z?

Answer 42

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

Answer 43

1 1 1 = - 2 (*-2)
2 -3 1 = -11



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

Answer 44

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 :)

Answer 45

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

Answer 46

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?

Answer 47

wait when did R1 become 1 2 -3 ?

Answer 48

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

Answer 49

oh ok

Answer 50

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

Answer 51

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?

Answer 52

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

Answer 53

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 :(

Answer 54

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

Answer 55

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

Answer 56

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

Answer 57

give me another

Answer 58

linear algebra has lots of matrix stuff in it :)

Answer 59

and maybe discrete mathematics

Answer 60

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

Answer 61

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

Answer 62

ok then ill do that