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anonymous
 3 years ago
<Solve the system>
a+c = 1
b+ac+d = 1
bc+ad = 1
bd = 1
anonymous
 3 years ago
<Solve the system> a+c = 1 b+ac+d = 1 bc+ad = 1 bd = 1

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bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Ok here goes... hopefully I won't make any arithmetic errors. From the first one: \[a+c= 1; \space c = 1a\]From the last one: \[b*d = 1; \space d = \frac{1}{b}\]2nd equation gives: \[b+a*c+d = 1\]\[b+a*(1a)+\frac{1}{b}= 1\]\[baa^2 + \frac{1}{b}=1\]3rd equation:\[b*c+a*d=1\]\[b*(1a)+\frac{a}{b}=1\]Let's see if we can solve 2nd and 3rd equations

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1\[b^2aba^2b+1=1\]\[b^2aba^2b=0\]\[b*(baa^2)=0\]

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.13rd equation:\[b^2ab^2+a=1\]

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Ok, now I'm stuck. @amistre64 @satellite73 , any ideas?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0try 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From 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 \[\frac{1}{d} + (1  c ) c + d =1\]\[\frac{1}{d} c + (1c) d = (1)\] How are you going to handle these equations?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0You 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It is easier said than done.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0From \(\frac{1}{d} + (1  c ) c + d =1\) \[\frac{1}{d} + (1  c ) c + d =1\]\[\frac{1}{d} + d =1+ (1 + c ) c\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{d} + d =1+ (1 + c ) c\]\[\frac{1}{d} + d =(c+\frac{1}{2})^2+\frac{3}{4}\]\[\frac{1}{d} + d \frac{3}{4}=(c+\frac{1}{2})^2\]\[(c+\frac{1}{2}) =\pm \sqrt{\frac{1}{d} + d \frac{3}{4}}\]\[c =\frac{1}{2}\pm \sqrt{\frac{1}{d} + d \frac{3}{4}}\]Wow!!~

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hummm... I cant find a way to make this a linear system and I don't know a good way to solve a nonlinear 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Class..... amateur :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@bahrom7893 The first line in your second comment here should be \[b^2−ab−a^2b+1=b\]
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