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anonymous

  • one year ago

Systems of Equations, please help!!

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  1. anonymous
    • one year ago
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    I know what the answers are but I cannot seem to get there =(

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  2. jim_thompson5910
    • one year ago
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    Let's focus on #24 \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2z = -3\\ z = 2x - 4y\\ \end{cases}\]

  3. jim_thompson5910
    • one year ago
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    Notice we have a pair of 'z's here \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2{\LARGE \color{red}{z}} = -3\\ {\LARGE \color{red}{z}} = 2x - 4y\\ \end{cases}\]

  4. jim_thompson5910
    • one year ago
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    what we can do is replace the 'z' in the second equation with '2x-4y' since z = 2x-4y in the third equation we can then drop the third equation after substitution \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{z} = -3\\ \color{red}{z} = 2x - 4y\end{cases}\] \[\Large \begin{cases} 13 = 3x - y\\ 4y-3x+2\color{red}{(2x-4y)} = -3\end{cases}\]

  5. anonymous
    • one year ago
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    For 24, which I think I maybe got right, I ended up with \[y=3x-13, 4y-3x+2z=-3, z=2x-4y\] and then substituted to end up with \[4(3x-13)-3x+2(2x-4(3x-13))=-3\] which simplified to \[x=5, y=2, z=2\]

  6. anonymous
    • one year ago
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    Is that correct (I have all the work but it's so long so I didn't post it sorry)

  7. jim_thompson5910
    • one year ago
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    You did the right steps. Nice job

  8. anonymous
    • one year ago
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    Awesome, thanks! Um do you think you could help with 25,26,27,28, or 29?

  9. jim_thompson5910
    • one year ago
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    Those are definitely harder because one variable isn't already isolated. But we can isolate a variable, say z if we pick on the third equation and solve for z, we get 3x - 2y - z = -9 3x - 2y - z+z = -9+z 3x - 2y = -9+z 3x - 2y + 9 = -9+z+9 3x - 2y + 9 = z z = 3x - 2y + 9

  10. jim_thompson5910
    • one year ago
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    agreed so far?

  11. anonymous
    • one year ago
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    For which question?

  12. jim_thompson5910
    • one year ago
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    #25

  13. anonymous
    • one year ago
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    Okay =) Yeah that looks good

  14. jim_thompson5910
    • one year ago
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    once we have z isolated, we can replace every copy of `z` in the first two equations with `(3x - 2y + 9)`

  15. anonymous
    • one year ago
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    Okay, did that =)

  16. jim_thompson5910
    • one year ago
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    first equation: `x+3y-z = -4` will turn into `x+3y-(3x - 2y + 9) = -4` second equation `2x-y+2z=13` will turn into `2x-y+2(3x - 2y + 9) =13`

  17. jim_thompson5910
    • one year ago
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    at this point, the 'z' has gone away leaving you with a system of 2 equations with 2 unknowns

  18. anonymous
    • one year ago
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    That makes sense! I think I tried to overcomplicate it last time lol. So from here, you would just isolate a variable in one of them and then plug whatever that is into the other equation? Or would you solve both for a variable and then solve it like a regular system?

  19. jim_thompson5910
    • one year ago
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    you can use a number of methods to solve this new system 1) substitution 2) elimination 3) matrices 4) graphing graphing is probably the fastest way, but it doesn't always guarantee you get the exact answers (the point may have fractional coordinates)

  20. anonymous
    • one year ago
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    I am most comfortable with substitution =) So I solve both for a variable (let's say y, so y=something) and then just substitute (if y=x and y=2 then 2=x)?

  21. jim_thompson5910
    • one year ago
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    you are correct

  22. jim_thompson5910
    • one year ago
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    let me know what you get

  23. anonymous
    • one year ago
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    Awesome thanks! Alright I'll do that real quick...

  24. anonymous
    • one year ago
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    I got y=2x-13 and y=(-5/3)

  25. jim_thompson5910
    • one year ago
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    x,y, & z are all whole numbers for #25

  26. anonymous
    • one year ago
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    True. And the answer for this is x=0, y=1, z=7... let's see what I did wrong.

  27. anonymous
    • one year ago
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    Ah. I combined x and y. I'll redo it

  28. jim_thompson5910
    • one year ago
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    how did you get `y=2x-13`? which equation did you solve for y?

  29. jim_thompson5910
    • one year ago
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    alright

  30. anonymous
    • one year ago
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    I solved the first one to get that.

  31. anonymous
    • one year ago
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    |dw:1442627727801:dw| Here's what I did to solve the first eq.

  32. jim_thompson5910
    • one year ago
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    You made a mistake in distributing x+3y-(3x - 2y + 9) = -4 turns into x+3y-3x + 2y - 9 = -4

  33. jim_thompson5910
    • one year ago
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    notice the `+ 2y` and ` - 9`

  34. anonymous
    • one year ago
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    OH! I was wondering about that. So the negative sign gets distributed!

  35. jim_thompson5910
    • one year ago
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    yeah think of `-(3x - 2y + 9)` as `-1*(3x - 2y + 9)`

  36. jim_thompson5910
    • one year ago
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    and that -1 multiplies with each term inside

  37. anonymous
    • one year ago
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    All right =) Here, I'll try that again

  38. anonymous
    • one year ago
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    I got y=(2/5x)+1

  39. anonymous
    • one year ago
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    and y=(8/5x)+1 for the other which when set equal to eachother make x=0 which is correct

  40. jim_thompson5910
    • one year ago
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    I'm getting the same equations when I solve for y

  41. anonymous
    • one year ago
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    Yay! When I plugged x=0 into the other eq.'s, it worked! Thank you so much!

  42. jim_thompson5910
    • one year ago
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    no problem

  43. anonymous
    • one year ago
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    I think I should be able to do the rest now. Thanks again =)

  44. jim_thompson5910
    • one year ago
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    sure thing

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