Solving Inequality and Number Line

J

jonboy

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Jun 8, 2006
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Hi guys I just want to make sure I did this problem right:

"Solve \(\displaystyle |\,3\,+\,6x\,|\,\ge\,6\) and graph the solution on a labeled number line."

So I have: \(\displaystyle \;3\,+\,6x\,\ge\,6\)

\(\displaystyle \L \;6x\,\ge\,3\,\;\Rightarrow\;x\,\ge\,\frac{1}{2}\)

or \(\displaystyle \;3\,+\,6x\,\le\,-\,6\)

\(\displaystyle \L \;x\,\le\,-\,\frac{3}{2}\)

So the answer would be no solution with a blank line?
 
Here is a simplification: \(\displaystyle \L\left| {3 + 6x} \right| \ge 6\quad \Leftrightarrow \quad \left| {1 + 2x} \right| \ge 2\)
 
OR not AND!

Try it out. Don't just guess.

x = 10

|1+2(10)| = |1+20| = |21| >= 2

I guess that kills the "no solution" theory.
 
In what follows assume \(\displaystyle a > 0\quad \& \quad b > 0\), then here are some general rules you might follow.

\(\displaystyle \left| {bx + c} \right| \ge a\quad \Rightarrow \quad \left\{ {\begin{array}{l}
{x \ge \frac{a - c }{b} } \\
{x \le \frac{{ - a - c }}{b} } \\
\end{array}} \right\}\quad \Rightarrow \quad \left( { - \infty ,\frac{{ - a - c }}{b} } \right] \cup \left[ {\frac{a-c}{b},\infty }\right).\)

\(\displaystyle \left| {bx + c} \right| \le a\quad \Rightarrow \quad \left\{ {\begin{array}{l}
{x \le \frac{a - c }{b}} \\
{x \ge \frac{{ - a - c }}{b}} \\
\end{array}} \right\}\quad \Rightarrow \quad \left[ {\frac{{ - a - c }}{b} ,\frac{a - c }{b} } \right].\)
 
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