How can you balance Cu+HNo3=Cu+(No3)2+No+H2o by using algebric or a b c d method please show me with full workout

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  1. well I am not familiar with your methods but maybe
    a Cu + b HNO3 = a Cu+c (NO3)2+ d NO + e H2O
    a = a obviously, Cu only appears once and a can be any number at all
    now H
    b = 2 e
    then N
    b = 2 c + d
    now O
    3 b = 6 c + d + e
    so now
    use 2 e for b
    2 e = 2 c + d
    3(2e) = 6 c + d + e or 5 e = 6 c + d
    so 3 e = 4 c
    e = (4/3) c
    then go back
    if c = 3 to get an integer number for e
    then e = 4
    d = 2 e - 2 c = 8 - 6 = 2
    and b = 2 e = 8
    try it:
    a Cu + 8 HNO3 = a Cu+ 3 (NO3)2+ 2 NO +4 H2O
    any old a
    8 H on each side
    8 N on each side
    24 O on each side
    It works

    I do not know why Cu, copper, is in your reaction since it does not react. Perhaps it is just to confuse you.

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  2. Two comments. First, for your trash kid, don't get excited about this. This is just an odd ball.
    Second. there is one BIG typo and several small ones. If you take care of the BIG one it takes care of Damon's problem AND you get the correct answer. Here is the corrected typo. I have removed the + sign between Cu and (NO3)2 on the product side. Now Damon doesn't have a problem because it shows Cu actually DID react. Now here is the correct equation using the caps key too.
    Cu + HNO3 ==> Cu(NO3)2 + NO + H2O
    Next you follow Damon's work and you end up with the balanced equation as follows:
    3Cu + 8HNO3 ==> 3Cu(NO3)2 + 2NO + 4H2O

    NOTE: This is a redox equation and I would NEVER use this method to balanced a reaction such as this one.
    Another note: Note that this reaction is for Cu and DILUTE HNO3. The product is NO + the others.
    When CONCENTRATED HNO3 is used the product is NO2 + others. That balanced equation is Cu + 4HNO3 ==> Cu(NO3)2 + 2NO2 + 2H2O.

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