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RolyPoly
<Solve the system> a+c = -1 b+ac+d = 1 bc+ad = -1 bd = 1
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
\[b^2-ab-a^2b+1=1\]\[b^2-ab-a^2b=0\]\[b*(b-a-a^2)=0\]
3rd equation:\[-b^2-ab^2+a=-1\]
Ok, now I'm stuck. @amistre64 @satellite73 , any ideas?
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.
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?
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.
It is easier said than done.
From \(\frac{1}{d} + (-1 - c ) c + d =1\) \[\frac{1}{d} + (-1 - c ) c + d =1\]\[\frac{1}{d} + d =1+ (1 + c ) c\]
\[\frac{1}{d} + d =1+ (1 + c ) c\]\[\frac{1}{d} + d =(c+\frac{1}{2})^2+\frac{3}{4}\]\[\frac{1}{d} + d -\frac{3}{4}=(c+\frac{1}{2})^2\]\[(c+\frac{1}{2}) =\pm \sqrt{\frac{1}{d} + d -\frac{3}{4}}\]\[c =-\frac{1}{2}\pm \sqrt{\frac{1}{d} + d -\frac{3}{4}}\]Wow!!~
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?
@bahrom7893 The first line in your second comment here should be \[b^2−ab−a^2b+1=b\]