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## Challenging System of Equations (4 Variables)

@HallsofIvy:
2/(a-1)^2+9b+9c-d=0 ... (1)
2/(b-1)^2+9a+9c-d=0 ... (2)
... doing (2)-(1) I get:
$$\begin{array}{rcccccccccl} \big (& 2/(b-1)^2 & +& 9a&+&9c&-&d&=&0& \big )\\ - \big (& 2/(a-1)^2 & + & 9b&+&9c&-&d&=&0 &\big ) \\ \hline = \bigg (& \frac{2}{(a-1)^2} - \frac{2}{(b-1)^2}& +&9(a-b)& & & & & =&0& \bigg ) \end{array}$$

How did you get rid of the 9(a-b) term?
Doesn't have a big effect on your argument - you still get f(a)=f(b) => a=b ?
Do the same with the other two and you get: a=b=c but d is different.
No need for shifting the axis about :)

But sometimes people have to go through the more painful ways first ;)

 Quote by Simon Bridge @HallsofIvy: 2/(a-1)^2+9b+9c-d=0 ... (1) 2/(b-1)^2+9a+9c-d=0 ... (2) ... doing (2)-(1) I get: $$\begin{array}{rcccccccccl} \big (& 2/(b-1)^2 & +& 9a&+&9c&-&d&=&0& \big )\\ - \big (& 2/(a-1)^2 & + & 9b&+&9c&-&d&=&0 &\big ) \\ \hline = \bigg (& \frac{2}{(a-1)^2} - \frac{2}{(b-1)^2}& +&9(a-b)& & & & & =&0& \bigg ) \end{array}$$ How did you get rid of the 9(a-b) term? Doesn't have a big effect on your argument - you still get f(a)=f(b) => a=b ? Do the same with the other two and you get: a=b=c but d is different. No need for shifting the axis about :) But sometimes people have to go through the more painful ways first ;)
So does that mean that if ##f(a)=f(b)=f(c)## then ##a=b=c## for every system of equations?

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 Quote by Chelonian So does that mean that if ##f(a)=f(b)=f(c)## then ##a=b=c## for every system of equations?
No - consider
##\text{if }f(x)=x^2,\text{ then } f(a)=f(b) \Rightarrow a=\pm b##
... so there are two possible values of a that satisfy the relation for a given value of b.
In this case, all the unknowns have to be positive.