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algebra: Given that a+c = 2, a+b = 3m^2, b+c = 6m, find the
- Thread starter tuzzi-i
- Start date
2. solve the following equation system
a^5 = b + b^5
b^5 = c + c^5
c^5 = d + d^5
d^5 = a + a^5
a=b=c=d=0 will work.
Re: algebra
Hello, tuzzi-i!
From [3] we have: \(\displaystyle \,c\:=\:2\,-\,a\)
Substitute into [2]: \(\displaystyle \,b\,+\,(2\,-\,a)\:=\:6m\)
. . and we have: \(\displaystyle \:-a\,+\,b\:=\:6m\,-\,2\)
. . . . . .Add [1]: \(\displaystyle \;\;a\,+\,b\:=\:3m^2\)
. . .Then: \(\displaystyle \,2b\:=\:3m^2\,+\,6m\,-\,2\;\;\Rightarrow\;\;b\:=\:\frac{3m^2\,+\,6m\,-\,2}{2}\)
Subtract: \(\displaystyle \,2a\:=\:3m^2\,-\,6m\,+\,2\;\;\Rightarrow\;\;a\:=\:\frac{3m^2\,-\,6m\,+\,2}{2}\)
From [3]: \(\displaystyle \frac{3m^2\,-\,6m\,+\,2}{2}\,+\,c\:=\:2\;\;\Rightarrow\;\;c\:=\:\frac{-3m^2\,+\,6m\,+\,2}{2}\)
Since \(\displaystyle a\,\leq\,b:\;\;\frac{3m^2\,-\,6m\,+\,2}{2}\:\leq\:\frac{3m^2\,+\,6m\,-\,2}{2}\;\;\Rightarrow\;\;m\:\geq\:\frac{1}{3}\)
Since \(\displaystyle b\,\leq\,c:\;\;\frac{3m^2\,+\,6m\,-\,2}{2}\:\leq\:\frac{-3m^2\,+\,6m\,+\,2}{2}\;\;\Rightarrow\;\;|m|\,\leq\,\sqrt{\frac{2}{3}}\)
Since \(\displaystyle a\,\leq\,c:\;\;\frac{3m^2\,-\.6m\,+\,2}{2}\:\leq\:\frac{-3m^2\,+\,6m\,+\,2}{2}\;\;\Rightarrow\;\;0\,\leq\,m\,\leq 2\)
Therefore: \(\displaystyle \:\frac{1}{3}\:\leq \:m\:\leq \:\sqrt{\frac{2}{3}}\)
Hello, tuzzi-i!
1. Given: \(\displaystyle \,\begin{array}{ccc}[1]\;a\,+\,b\:=\:3m^2 \\ [2]\;b\,+\,c\:=\:6m\; \\ [3]\;a\,+\,c\:=\:2\;\;\,\end{array}\;\)where \(\displaystyle m\) is a real number
Find the values of \(\displaystyle m\) if \(\displaystyle a\,\leq\,b\leq\,c\)
From [3] we have: \(\displaystyle \,c\:=\:2\,-\,a\)
Substitute into [2]: \(\displaystyle \,b\,+\,(2\,-\,a)\:=\:6m\)
. . and we have: \(\displaystyle \:-a\,+\,b\:=\:6m\,-\,2\)
. . . . . .Add [1]: \(\displaystyle \;\;a\,+\,b\:=\:3m^2\)
. . .Then: \(\displaystyle \,2b\:=\:3m^2\,+\,6m\,-\,2\;\;\Rightarrow\;\;b\:=\:\frac{3m^2\,+\,6m\,-\,2}{2}\)
Subtract: \(\displaystyle \,2a\:=\:3m^2\,-\,6m\,+\,2\;\;\Rightarrow\;\;a\:=\:\frac{3m^2\,-\,6m\,+\,2}{2}\)
From [3]: \(\displaystyle \frac{3m^2\,-\,6m\,+\,2}{2}\,+\,c\:=\:2\;\;\Rightarrow\;\;c\:=\:\frac{-3m^2\,+\,6m\,+\,2}{2}\)
Since \(\displaystyle a\,\leq\,b:\;\;\frac{3m^2\,-\,6m\,+\,2}{2}\:\leq\:\frac{3m^2\,+\,6m\,-\,2}{2}\;\;\Rightarrow\;\;m\:\geq\:\frac{1}{3}\)
Since \(\displaystyle b\,\leq\,c:\;\;\frac{3m^2\,+\,6m\,-\,2}{2}\:\leq\:\frac{-3m^2\,+\,6m\,+\,2}{2}\;\;\Rightarrow\;\;|m|\,\leq\,\sqrt{\frac{2}{3}}\)
Since \(\displaystyle a\,\leq\,c:\;\;\frac{3m^2\,-\.6m\,+\,2}{2}\:\leq\:\frac{-3m^2\,+\,6m\,+\,2}{2}\;\;\Rightarrow\;\;0\,\leq\,m\,\leq 2\)
Therefore: \(\displaystyle \:\frac{1}{3}\:\leq \:m\:\leq \:\sqrt{\frac{2}{3}}\)