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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
 one year ago
RolyPoly Group Title
<Solve the system> a+c = 1 b+ac+d = 1 bc+ad = 1 bd = 1
 one year ago
 one year ago

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bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
Ok 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
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
\[b^2aba^2b+1=1\]\[b^2aba^2b=0\]\[b*(baa^2)=0\]
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
3rd equation:\[b^2ab^2+a=1\]
 one year ago

bahrom7893 Group TitleBest ResponseYou've already chosen the best response.0
Ok, now I'm stuck. @amistre64 @satellite73 , any ideas?
 one year ago

JBrandonS Group TitleBest ResponseYou've already chosen the best response.0
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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
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 \[\frac{1}{d} + (1  c ) c + d =1\]\[\frac{1}{d} c + (1c) d = (1)\] How are you going to handle these equations?
 one year ago

JBrandonS Group TitleBest ResponseYou've already chosen the best response.0
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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
It is easier said than done.
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{1}{d} + d =1+ (1 + c ) c\]\[\frac{1}{d} + d =(c+\frac{1}{2})^2+\frac{3}{4}\]\[\frac{1}{d} + d \frac{3}{4}=(c+\frac{1}{2})^2\]\[(c+\frac{1}{2}) =\pm \sqrt{\frac{1}{d} + d \frac{3}{4}}\]\[c =\frac{1}{2}\pm \sqrt{\frac{1}{d} + d \frac{3}{4}}\]Wow!!~
 one year ago

JBrandonS Group TitleBest ResponseYou've already chosen the best response.0
hummm... 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?
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Class..... amateur :D
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
@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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