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anonymous

  • 5 years ago

how does 3+2(-11/14) become 3-11/7?

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  1. anonymous
    • 5 years ago
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    14 is cancelled by table 2 as 2*7=14

  2. xkehaulanix
    • 5 years ago
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    According to PEMDAS, the order in which you do operations is such: Parenthesis Exponents Multiplication Division Addition Subtraction Since you have multiplication in the problem, you would do the multiplication before the addition. 3 + 2 x (-11/14) The 2 cancels part of the 14 because it is factorable by 2: 2 x -11/(2 x 7) = -11/7 Since the 3 is unaffected by the multiplication, it remains in the equation.

  3. anonymous
    • 5 years ago
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    I think I understand.

  4. anonymous
    • 5 years ago
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    thats gud.

  5. anonymous
    • 5 years ago
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    so how would you proceed to the next step? x=(3-11/7)-2z.......How do you solve 3-11/7

  6. anonymous
    • 5 years ago
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    to solve 3-11/7 taking L.C.M, =(21-11)/7 =11/7

  7. anonymous
    • 5 years ago
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    sory its 10/7

  8. anonymous
    • 5 years ago
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    l.c.m?

  9. anonymous
    • 5 years ago
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    so does 3-11/7 become 3/1*11/7 and cross multiply 3and 7, 1 and 11?

  10. anonymous
    • 5 years ago
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    but then wouldn't the answer be 3?

  11. anonymous
    • 5 years ago
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    yes l.c.m

  12. anonymous
    • 5 years ago
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    l.c.m is of the numbers in denominatot

  13. anonymous
    • 5 years ago
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    3/1-11/7 so we have 1 and 7 in denominator so 7 is the l.c.m

  14. anonymous
    • 5 years ago
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    no its not cross multiplication i just took l.c.m

  15. xkehaulanix
    • 5 years ago
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    l.c.m stands for "Least Common Multiple," but you can call it the least common denominator too. To add 3 to 11/7, you need to make sure that the denominators are the same. So yes, 3 is currently divided by 1, and 11 is divided by 7. You can't cross multiply for this part though because the numbers are not in a proportion. So to make the denominator on 3 equal to the denominator under 11, you would need to multiply it by 7. So 3/1 x 7/7= 21/7

  16. anonymous
    • 5 years ago
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    I'm sorry, I don't even understand what LCM is, I haven't done math in 15 years

  17. anonymous
    • 5 years ago
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    i got it what are u sayng,its ok in ur sense cross multiply after l.c.m

  18. anonymous
    • 5 years ago
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    ok dnt worry i tell what the l.c.m is

  19. anonymous
    • 5 years ago
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    so then it becomes 21/7-11/7?

  20. anonymous
    • 5 years ago
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    and then 10/7 okay!

  21. xkehaulanix
    • 5 years ago
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    Yup! That's it!

  22. anonymous
    • 5 years ago
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    so now what is the next step, x=10/7-2z

  23. anonymous
    • 5 years ago
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    as on present case we have 1,7 so we divide both on a single number until we get 1 then multiply those numbers which is l.c.m

  24. xkehaulanix
    • 5 years ago
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    Are you solving for z?

  25. anonymous
    • 5 years ago
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    ok its 2z?solving 4 z or x?

  26. anonymous
    • 5 years ago
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    ummmm, this was the original equation, 4x+6y-8z=1 -6x-9y+12z=0 x-2y-2z=3

  27. xkehaulanix
    • 5 years ago
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    Ohh. You're supposed to solve for all of the variables, right?

  28. anonymous
    • 5 years ago
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    ok u have system of linear equations to solve

  29. anonymous
    • 5 years ago
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    yes, w/ help I got this far, x=3-2y-2z 4(3-2y-2z)+6y-8z=1 -6(3-2y-2z)-9y+12z=0 and then 4(3-2y-2z)+6y-8z=1 12+8y+8z+6y-8z=1 12+14y=1 14y=-11 y=-11/14 so x=3+2(-11/14)-2z x=(3-11/7)-2z and so now I'm at x=10/7-2z

  30. anonymous
    • 5 years ago
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    so now I know y=10/7 or 1 3/7?

  31. xkehaulanix
    • 5 years ago
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    Wow..that looks like a lot to type. Hmm, ok. There's a small error in the beginning. The third equation you wrote was x-2y-2z=3, right? In the last post you made, you replaced x with 3-2y-2z. Those equations are actually not equal, but it's just a problem with positive and negative signs. x-2y-2z=3 Since you want x separated from the other two variables, you would move them to the other side of the equation. First, you would add 2y to both sides to keep the equation balanced (remember, what you do to one side of an equation must be done to the other, just like a seesaw). So add 2y and 2z to both sides to separate x. x=3+2y+2z

  32. anonymous
    • 5 years ago
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    okay, so then it becomes 4(3+2y+2z) +6y-8z=1 12+8y+8z+6y-8z=1 12+14y=1.....so it still comes out the same in the end doesn't it? Am I not still left with x=10/7-2z?

  33. anonymous
    • 5 years ago
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    OOO no, it changes it to x=10/7+2z

  34. xkehaulanix
    • 5 years ago
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    Whoa, you lost me xD Well from there you can get y, which is still correct. And if you plug y into that equation, it becomes x=3+2(-11/14)+2z So yes, the last one you posted was right.

  35. anonymous
    • 5 years ago
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    okay, so how do I solve it? how do I solve x=10/7+2z?

  36. anonymous
    • 5 years ago
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    do I now know that y=10/7? how do I figure out x and z?

  37. xkehaulanix
    • 5 years ago
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    Okay, so far you've only had to use two equations. You can plug in the new x value (10/7 + 2z) into the last equation to solve for z since you already know that y is equal to -11/14. The y you got from the 12+14y=1 that you solved earlier. So you already know what y is!

  38. anonymous
    • 5 years ago
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    so now the equation is -6(10/7+2z)-9(10/7)+12z=0?

  39. xkehaulanix
    • 5 years ago
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    Err...yes to everything except the 10/7. The equation for y you had earlier was 12+14y=1 When you solved for it, you ended up with -11/14.

  40. anonymous
    • 5 years ago
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    oh....ok so 10/7 only works in the x-2y-2z=3 equation?

  41. anonymous
    • 5 years ago
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    or the 4x+6y-8z=1 equation?

  42. anonymous
    • 5 years ago
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    I thought I was getting it, but now I'm confused all over again

  43. xkehaulanix
    • 5 years ago
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    The 10/7 you got in that equation wasn't equal to any of the variables, actually. It was just part of the equation for x. The equation was x=10/7 + 2z, right? All the 10/7 tells you is that x is 10/7 larger than twice of z. It doesn't stand for a variable. xD Math is such a joy, isn't it?

  44. anonymous
    • 5 years ago
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    lol yea

  45. anonymous
    • 5 years ago
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    okay. so y=-11/14

  46. anonymous
    • 5 years ago
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    so now I do x=2(-11/7)+2z?

  47. anonymous
    • 5 years ago
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    or 4(3+2(-11/7)+2z)+6(-11/14)-8z=1 and -6(3+2(-11/14)+2z)-9(-11/14)+12z=0?

  48. anonymous
    • 5 years ago
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    becoming 12+8(-11/14)+6(-11/14)+4z=1and -18-12(-11/14)-9(-11/14)=0

  49. xkehaulanix
    • 5 years ago
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    Whoops, sorry. I had to go find the equations again. Whoaaah. You could do that, but trying to keep your variables straight gets really confusing, doesn't it? When it comes to multiple equations like this, it's usually easier to try and separate a variable before putting in more values. So instead of plugging in the values for x and y in the beginning, try separating x or z from one of the equations first. Say we took 4x+6y-8z=1 We could pull out the z to make it a little less confusing. So 4x+6y=1+8z 4x+6y-1=8z And you could divide both sides by 8 if you really wanted to, but it depends on how comfortable you are with fractions...and on a computer screen, fractions are hard to keep track of. Then for y you would plug in -11/14

  50. anonymous
    • 5 years ago
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    I'm good with fractions for the most part, it's the rest of this junk

  51. anonymous
    • 5 years ago
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    and I'm keeping track on paper too :0)

  52. anonymous
    • 5 years ago
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    so it would become 4x+6(-11/14) -1=8z?

  53. xkehaulanix
    • 5 years ago
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    And for x, you could plug in either the original equation of x=3+2y+2z or the simpler x=10/7+2z that you figured out earlier. Oh, then you shouldn't have any problems xD I think I'm avoiding fractions for myself, just because I never see the division sign...

  54. xkehaulanix
    • 5 years ago
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    Yup. It has a lot less numbers than before, right? xD

  55. anonymous
    • 5 years ago
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    okay....so 4(10/7+2z) +6(-11/14)-1=8z

  56. anonymous
    • 5 years ago
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    yea it does

  57. anonymous
    • 5 years ago
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    okay, so I figured the first half, 40+8z, how do you multiple a positive with a negative?6(-11/14)

  58. xkehaulanix
    • 5 years ago
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    To multiply a negative number, the rules are pretty simple. If you multiply a negative by a positive (as in this case), the result will be a negative number. It multiplies the same as two positives though. If you have two negative numbers, then the result will be positive instead.

  59. anonymous
    • 5 years ago
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    aha, so then 6(-11/14) becomes -66

  60. anonymous
    • 5 years ago
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    so then the equation becomes 2z-27=8z

  61. xkehaulanix
    • 5 years ago
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    Yep, but don't forget to divide by the 14. OH NO. I'm sorry, I wasn't paying attention to the equation I picked! We already used this equation to solve for the x value, so the variables are just going to cancel out...You have to use the third equation to solve for z. So use -6x-9y+12z=0... Ahh...my bad.

  62. anonymous
    • 5 years ago
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    oh okay, hold on. let me try

  63. anonymous
    • 5 years ago
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    okay, so: -6x-9y+12z=0 -6x-9y=-12z -6(10/7)+2z)-9(-11/14)=-12z -60+2z+99=-12z 39+2z=-12z 39=-12z-2z 39=-14z?

  64. anonymous
    • 5 years ago
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    so z=2 -11/14

  65. anonymous
    • 5 years ago
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    or more like -2 11/14

  66. xkehaulanix
    • 5 years ago
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    Ahh...almost. You have the equation right, but somewhere in the middle of multiplying you dropped your denominators. Take 6[(10/7)+2z], for example. When you multiply by the 6, you should end up with (60/7)+12z, rather than 60+12z.

  67. anonymous
    • 5 years ago
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    ummmm....6(10/7) ooooohhh okay

  68. xkehaulanix
    • 5 years ago
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    Mmhmm. The same goes for the -9(-11/14). You got the positive sign right on that part though :D

  69. anonymous
    • 5 years ago
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    okay, i'll try again

  70. anonymous
    • 5 years ago
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    so -6 is a whole number, so to multiply it by 10/7 you have to make it a fraction correct? so it becomes -6/1x10/7. then you have to have common denominators and what you do to the bottom you do to the top? -42/7x10/7 then becomes -420/7 becoming -60?

  71. anonymous
    • 5 years ago
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    or was that other person wrong, and it's just -6/1x10/7=-60/7?

  72. anonymous
    • 5 years ago
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    -8 4/7(simplified)

  73. xkehaulanix
    • 5 years ago
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    The least common denominator is really only used when adding and subtracting. So the other person was right on that part. For multiplication, the last thing you posted is correct. I'd be careful with mixed fractions (the whole number with the fraction) only because multiplying them gets tricky. And...I think something went wrong again because my variable disappeared. Let me try doing the problem on paper ^-^'

  74. anonymous
    • 5 years ago
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    okay, i'm working it out on paper too, and this is where i'm at, -60/7+2z-99/14=-12z -219/14+2z=-12z -219/14=-12z-2z -219/14=-14z

  75. anonymous
    • 5 years ago
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    so now I divide -219/14 by -14 to =z?

  76. anonymous
    • 5 years ago
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    -219/14 divided by-14/1=-219/14x-1/14=219/196

  77. xkehaulanix
    • 5 years ago
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    There's one more thing missing in your multiplication. When you multiply something inside of parenthesis, you have to multiply everything inside. So you had 6[(10/7)+2z], right? You have to multiply both the 10/7 AND the 2z by 6 because they're inside the parenthesis. It goes back to the PEMDAS thing I mentioned a long time ago.

  78. anonymous
    • 5 years ago
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    crap. I thought I did that.

  79. anonymous
    • 5 years ago
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    okay so -60/7-12z+99/14=-12z

  80. anonymous
    • 5 years ago
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    therefore, -3/2-12z=-12z

  81. xkehaulanix
    • 5 years ago
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    Yup, that's what I've been getting. But from there, the z keep getting dropped, no matter which equation I use..

  82. anonymous
    • 5 years ago
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    -3/2=-12z+12z

  83. anonymous
    • 5 years ago
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    yea, it did it for me too, so does that mean there is no solution? we find out x and y but never z?

  84. xkehaulanix
    • 5 years ago
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    Technically, you can find z if you find x. But we shouldn't be dropping z in every equation we try because...well, it's not really logical, I guess. I'm still looking to see if I messed up too.

  85. anonymous
    • 5 years ago
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    okay

  86. xkehaulanix
    • 5 years ago
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    Ahh, no good. I'm running in circles now. Are you sure the original equations are right?

  87. anonymous
    • 5 years ago
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    That's what I was given. :0( Well thank you for all your help, I think it's time to call it a night. I will have to double check the equation tomorrow, and try again if the originals were wrong. Thank you for your patience, at least I know how to figure it out next time!

  88. xkehaulanix
    • 5 years ago
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    Okay then. Sorry I couldn't help more though!

  89. anonymous
    • 5 years ago
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    It's okay, you were great! Thank you!

  90. xkehaulanix
    • 5 years ago
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    No problem! Good luck tomorrow!

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