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anonymous

  • 5 years ago

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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  1. amistre64
    • 5 years ago
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    [4 -7 | -2] [1 2 | 7]

  2. amistre64
    • 5 years ago
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    matrix is just the elimination process without the variables getting in the way

  3. amistre64
    • 5 years ago
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    [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

  4. amistre64
    • 5 years ago
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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

  5. amistre64
    • 5 years ago
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    [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]

  6. amistre64
    • 5 years ago
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    with any luck; x=3 and y = 2

  7. amistre64
    • 5 years ago
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    4(3)-7(2)=-2 12 - 14 = -2 checks (3)+2(2)=7 3 + 4 = 7 checks

  8. amistre64
    • 5 years ago
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    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

  9. anonymous
    • 5 years ago
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    dear god thank you tho i still dnt quite understand how to do the other 4 :(

  10. amistre64
    • 5 years ago
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    post them lol

  11. anonymous
    • 5 years ago
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    k thank you

  12. amistre64
    • 5 years ago
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    instead of trying to hunt down the [,|, and ] buttons im gonna leave them out ot it ;)

  13. anonymous
    • 5 years ago
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    alrighty thanks want me to post here or under ask a question?

  14. amistre64
    • 5 years ago
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    youre call ;)

  15. anonymous
    • 5 years ago
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    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

  16. anonymous
    • 5 years ago
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    ill post here

  17. amistre64
    • 5 years ago
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    :)

  18. anonymous
    • 5 years ago
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    11. 5x=3y-50 2y=1-3x 13. x+y+z=-2 2x-3y+z=-11 -x+2y-z=8

  19. amistre64
    • 5 years ago
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    do you know how to solve a system of equations by elimination?

  20. anonymous
    • 5 years ago
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    if i saw the problem i could tell you but i dnt remember alot of precal vocab

  21. amistre64
    • 5 years ago
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    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

  22. amistre64
    • 5 years ago
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    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

  23. anonymous
    • 5 years ago
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    the ansers shld come out to be x= -97/19 and y= 155/19 in the back of the book

  24. amistre64
    • 5 years ago
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    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 :)

  25. amistre64
    • 5 years ago
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    let me recheck without the explanations ok....

  26. amistre64
    • 5 years ago
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    5x -3y=-50 3x+2y = 1 <- had a -3 there...bad me lol

  27. anonymous
    • 5 years ago
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    its fine man

  28. amistre64
    • 5 years ago
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    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 ??

  29. amistre64
    • 5 years ago
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    31 5 --- 155 lol 155/19

  30. amistre64
    • 5 years ago
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    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?

  31. amistre64
    • 5 years ago
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    its easier on the paper lol

  32. amistre64
    • 5 years ago
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    does it look any easier yet :)

  33. anonymous
    • 5 years ago
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    starting to but i look at these problems and i just feel myself getting lost :(

  34. amistre64
    • 5 years ago
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    1 1 1 = - 2 2 -3 1 = -11 -1 2 -1 = 8

  35. amistre64
    • 5 years ago
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    when we look at this the 1st and 3rd are ready to be added togeter and eliminate that -1 right

  36. anonymous
    • 5 years ago
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    oh the first and third 1 tho

  37. anonymous
    • 5 years ago
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    dnt we need the bottom right to be 1?

  38. anonymous
    • 5 years ago
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    i guess we do that later?

  39. amistre64
    • 5 years ago
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    1 1 1 = - 2 -1 2 -1 = 8 ------------ 0 3 0 = 6 (/3) 0 1 0 = 2 ..... y = 2

  40. amistre64
    • 5 years ago
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    we use R1 to calibrate R2 and R3

  41. anonymous
    • 5 years ago
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    but hte third row is z?

  42. anonymous
    • 5 years ago
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    so dnst it also need it to be 0 0 1?

  43. amistre64
    • 5 years ago
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    1 1 1 = - 2 (*-2) 2 -3 1 = -11 -2 -2 -2= 4 2 -3 1 = -11 -------------- 0 -5 -1 = -7 <- new R3

  44. amistre64
    • 5 years ago
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    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 :)

  45. amistre64
    • 5 years ago
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    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

  46. amistre64
    • 5 years ago
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    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?

  47. anonymous
    • 5 years ago
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    wait when did R1 become 1 2 -3 ?

  48. amistre64
    • 5 years ago
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    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

  49. anonymous
    • 5 years ago
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    oh ok

  50. amistre64
    • 5 years ago
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    whenever we know what a '1' in a column equals; we can substitute it to find aonther :)

  51. anonymous
    • 5 years ago
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    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?

  52. amistre64
    • 5 years ago
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    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

  53. anonymous
    • 5 years ago
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    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 :(

  54. amistre64
    • 5 years ago
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    x y z = # --------- 1 0 0 = -1 0 1 0 = 2 0 0 1 = -3

  55. anonymous
    • 5 years ago
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    i dont even get the option of failing this cause my navy nuke recruiter wld kill me :(

  56. amistre64
    • 5 years ago
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    lol.... i dont know much about matrixes, but i do know this :)

  57. amistre64
    • 5 years ago
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    give me another

  58. amistre64
    • 5 years ago
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    linear algebra has lots of matrix stuff in it :)

  59. amistre64
    • 5 years ago
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    and maybe discrete mathematics

  60. amistre64
    • 5 years ago
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    post a new question box so we dont bog this post down ...long posts tend to slow my system ;)

  61. anonymous
    • 5 years ago
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    k ill throw up the last two of these and grab a snack and brb here they are

  62. anonymous
    • 5 years ago
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    ok then ill do that

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