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

Mathematics
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[4 -7 | -2] [1 2 | 7]
matrix is just the elimination process without the variables getting in the way
[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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[ 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
[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]
with any luck; x=3 and y = 2
4(3)-7(2)=-2 12 - 14 = -2 checks (3)+2(2)=7 3 + 4 = 7 checks
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
dear god thank you tho i still dnt quite understand how to do the other 4 :(
post them lol
k thank you
instead of trying to hunt down the [,|, and ] buttons im gonna leave them out ot it ;)
alrighty thanks want me to post here or under ask a question?
youre call ;)
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
ill post here
:)
11. 5x=3y-50 2y=1-3x 13. x+y+z=-2 2x-3y+z=-11 -x+2y-z=8
do you know how to solve a system of equations by elimination?
if i saw the problem i could tell you but i dnt remember alot of precal vocab
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
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
the ansers shld come out to be x= -97/19 and y= 155/19 in the back of the book
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 :)
let me recheck without the explanations ok....
5x -3y=-50 3x+2y = 1 <- had a -3 there...bad me lol
its fine man
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 ??
31 5 --- 155 lol 155/19
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?
its easier on the paper lol
does it look any easier yet :)
starting to but i look at these problems and i just feel myself getting lost :(
1 1 1 = - 2 2 -3 1 = -11 -1 2 -1 = 8
when we look at this the 1st and 3rd are ready to be added togeter and eliminate that -1 right
oh the first and third 1 tho
dnt we need the bottom right to be 1?
i guess we do that later?
1 1 1 = - 2 -1 2 -1 = 8 ------------ 0 3 0 = 6 (/3) 0 1 0 = 2 ..... y = 2
we use R1 to calibrate R2 and R3
but hte third row is z?
so dnst it also need it to be 0 0 1?
1 1 1 = - 2 (*-2) 2 -3 1 = -11 -2 -2 -2= 4 2 -3 1 = -11 -------------- 0 -5 -1 = -7 <- new R3
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 :)
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
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?
wait when did R1 become 1 2 -3 ?
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
oh ok
whenever we know what a '1' in a column equals; we can substitute it to find aonther :)
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?
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
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 :(
x y z = # --------- 1 0 0 = -1 0 1 0 = 2 0 0 1 = -3
i dont even get the option of failing this cause my navy nuke recruiter wld kill me :(
lol.... i dont know much about matrixes, but i do know this :)
give me another
linear algebra has lots of matrix stuff in it :)
and maybe discrete mathematics
post a new question box so we dont bog this post down ...long posts tend to slow my system ;)
k ill throw up the last two of these and grab a snack and brb here they are
ok then ill do that

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