i am stumped on this problem can someone help......
- Thread starter ejames2
- Start date
Hi, ejames2!
\(\displaystyle \L \;\frac{2\,+\,x}{2\,-\,x}\,\le\,1\)
Multiply by \(\displaystyle 2\,-\,x\):\(\displaystyle \L \;2\,+x\,\le\,2\,-\,x\)
Subtract \(\displaystyle 2\) and add \(\displaystyle x\):\(\displaystyle \L \;2x\,\le\,0\,\Rightarrow\,x\,\le\,0\)
Check with \(\displaystyle \,-\,1\):\(\displaystyle \L \;\frac{2\,-\,1}{2\,+\,1}\,\le\,1\)
\(\displaystyle \L \;\frac{1}{3}\,\le\,1\)
\(\displaystyle \L \;\frac{2\,+\,x}{2\,-\,x}\,\le\,1\)
Multiply by \(\displaystyle 2\,-\,x\):\(\displaystyle \L \;2\,+x\,\le\,2\,-\,x\)
Subtract \(\displaystyle 2\) and add \(\displaystyle x\):\(\displaystyle \L \;2x\,\le\,0\,\Rightarrow\,x\,\le\,0\)
Check with \(\displaystyle \,-\,1\):\(\displaystyle \L \;\frac{2\,-\,1}{2\,+\,1}\,\le\,1\)
\(\displaystyle \L \;\frac{1}{3}\,\le\,1\)
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,990
Jonboy, there is a problem with your solution.
It is not complete because you cannot multiply unless you know the term is positive.
\(\displaystyle \L
\begin{array}{l}
\frac{{2 + x}}{{2 - x}} \le 1 \\
\frac{{2 + x}}{{2 - x}} - 1 \le 0 \\
\frac{{2x}}{{2 - x}} \le 0\quad \Rightarrow \quad x \le 0\quad \cup \quad 2 < x \\
\end{array}\)
It is not complete because you cannot multiply unless you know the term is positive.
\(\displaystyle \L
\begin{array}{l}
\frac{{2 + x}}{{2 - x}} \le 1 \\
\frac{{2 + x}}{{2 - x}} - 1 \le 0 \\
\frac{{2x}}{{2 - x}} \le 0\quad \Rightarrow \quad x \le 0\quad \cup \quad 2 < x \\
\end{array}\)