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RolyPoly Group Title

<Solve the system> a+c = -1 b+ac+d = 1 bc+ad = -1 bd = 1

  • one year ago
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  1. bahrom7893 Group Title
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    Ok here goes... hopefully I won't make any arithmetic errors. From the first one: \[a+c= -1; \space c = -1-a\]From the last one: \[b*d = 1; \space d = \frac{1}{b}\]2nd equation gives: \[b+a*c+d = 1\]\[b+a*(-1-a)+\frac{1}{b}= 1\]\[b-a-a^2 + \frac{1}{b}=1\]3rd equation:\[b*c+a*d=-1\]\[b*(-1-a)+\frac{a}{b}=-1\]Let's see if we can solve 2nd and 3rd equations

    • one year ago
  2. bahrom7893 Group Title
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    \[b^2-ab-a^2b+1=1\]\[b^2-ab-a^2b=0\]\[b*(b-a-a^2)=0\]

    • one year ago
  3. bahrom7893 Group Title
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    3rd equation:\[-b^2-ab^2+a=-1\]

    • one year ago
  4. bahrom7893 Group Title
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    Ok, now I'm stuck. @amistre64 @satellite73 , any ideas?

    • one year ago
  5. JBrandonS Group Title
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    try using the last equation to get d in terms of b, and the first equation to get c in terms of a. this should let you plug into the second and third equations to have a system of 2 variables and 2 functions. solve that and then use those value to solve the first and last equation for d and b.

    • one year ago
  6. RolyPoly Group Title
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    From a+c = -1, we get a = -1 - c ---(1) From bd = 1, we get b= 1/d ---(2) Sub (1) and (2) into b+ac+d = 1 and bc+ad = -1 \[\frac{1}{d} + (-1 - c ) c + d =1\]\[\frac{1}{d} c + (-1-c) d = (-1)\] How are you going to handle these equations?

    • one year ago
  7. JBrandonS Group Title
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    You can pick one equation, solve for either c or d and substitute into the other. From there you will be able to solve for a variable. Then you plug into the first equation you picked to get the second variable. Then you can solve both (1) and (2) easy.

    • one year ago
  8. RolyPoly Group Title
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    It is easier said than done.

    • one year ago
  9. RolyPoly Group Title
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    From \(\frac{1}{d} + (-1 - c ) c + d =1\) \[\frac{1}{d} + (-1 - c ) c + d =1\]\[\frac{1}{d} + d =1+ (1 + c ) c\]

    • one year ago
  10. RolyPoly Group Title
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    \[\frac{1}{d} + d =1+ (1 + c ) c\]\[\frac{1}{d} + d =(c+\frac{1}{2})^2+\frac{3}{4}\]\[\frac{1}{d} + d -\frac{3}{4}=(c+\frac{1}{2})^2\]\[(c+\frac{1}{2}) =\pm \sqrt{\frac{1}{d} + d -\frac{3}{4}}\]\[c =-\frac{1}{2}\pm \sqrt{\frac{1}{d} + d -\frac{3}{4}}\]Wow!!~

    • one year ago
  11. JBrandonS Group Title
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    hummm... I cant find a way to make this a linear system and I don't know a good way to solve a non-linear system that I would want to explain. Wolfram alpha is coming up with some very ugly numbers for this one too so I don't think this is a easy one to solve. What class is this for?

    • one year ago
  12. RolyPoly Group Title
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    Class..... amateur :D

    • one year ago
  13. RolyPoly Group Title
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    @bahrom7893 The first line in your second comment here should be \[b^2−ab−a^2b+1=b\]

    • one year ago
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