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

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

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

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

JBrandonS
 one year 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.

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1From 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?

JBrandonS
 one year 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.

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

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1From \(\frac{1}{d} + (1  c ) c + d =1\) \[\frac{1}{d} + (1  c ) c + d =1\]\[\frac{1}{d} + d =1+ (1 + c ) c\]

RolyPoly
 one year ago
Best 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!!~

JBrandonS
 one year 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?

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

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