MHB Number Line & Intervals (Part 2)

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The discussion focuses on solving inequalities involving absolute values and representing the solutions on a number line. For the inequality |x - 4| < 4, the solution is the interval (0, 8), indicating all real numbers between 0 and 8. For the inequality |x + 5| >= 2, the solution consists of two intervals: (-∞, -7] and [-3, ∞), representing all numbers whose distance from -5 is greater than or equal to 2. Participants confirm the accuracy of these intervals and provide visual representations on a number line. The conversation emphasizes understanding absolute value inequalities and their graphical interpretations.
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The set of real numbers satisfying the given inequality is one or more intervals on the number line. Show the intervals on a number line.

(A) |x - 4| < 4

(B) |x + 5| >= 2

For (A), I did the following:

-4 < x - 4 < 4

I now add 4 to each term.

0 < x < 8

On the number line, I would need to plot (0, 8). Is this correct?

For (B), we have the following:

|x + 5| >= 2

x + 5 < -2 or x + 5 >= 2

x =< -2 - 5 or x >= 2 - 5

x =< - 7 or x >= -3

I must plot [-infinity, -7] and [-3, infinity] on the number line. Is this right?
 
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a) $$|x-4|<4$$

I would read this as a distance formula, that is, all real numbers $x$ that are less than 4 units from 4, which as you stated is the interval $(0,8)$.

\begin{tikzpicture}[scale=2.5]
\draw[very thick] (0,0) -- (8,0);
\path [draw=black, fill=white, thick] (0,0) circle (2pt);
\path [draw=black, fill=white, thick] (8,0) circle (2pt);
\draw[latex-latex] (-0.5,0) -- (8.5,0) ;
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

b) $$|x+5|\ge2$$

I would read this as all real numbers $x$ whose distance from $-5$ is greater than or equal to 2, which as you stated is the interval $(-\infty,-7]\,\cup\,[-3,\infty)$.

\begin{tikzpicture}[scale=2.5]
\path [draw=black, fill=black, thick] (-7,0) circle (2pt);
\path [draw=black, fill=black, thick] (-3,0) circle (2pt);
\draw[latex-latex] (-9.5,0) -- (-0.5,0) ;
\draw[->,thick] (-7,0) -- (-9.25,0);
\draw[->,thick] (-3,0) -- (-0.75,0);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}
 
MarkFL said:
a) $$|x-4|<4$$

I would read this as a distance formula, that is, all real numbers $x$ that are less than 4 units from 4, which as you stated is the interval $(0,8)$.

\begin{tikzpicture}[scale=2.5]
\draw[very thick] (0,0) -- (8,0);
\path [draw=black, fill=white, thick] (0,0) circle (2pt);
\path [draw=black, fill=white, thick] (8,0) circle (2pt);
\draw[latex-latex] (-0.5,0) -- (8.5,0) ;
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {0,1,2,3,4,5,6,7,8}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

b) $$|x+5|\ge2$$

I would read this as all real numbers $x$ whose distance from $-5$ is greater than or equal to 2, which as you stated is the interval $(-\infty,-7]\,\cup\,[-3,\infty)$.

\begin{tikzpicture}[scale=2.5]
\path [draw=black, fill=black, thick] (-7,0) circle (2pt);
\path [draw=black, fill=black, thick] (-3,0) circle (2pt);
\draw[latex-latex] (-9.5,0) -- (-0.5,0) ;
\draw[->,thick] (-7,0) -- (-9.25,0);
\draw[->,thick] (-3,0) -- (-0.75,0);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,3pt) -- (0pt,-3pt);
\foreach \x in {-9,-8,-7,-6,-5,-4,-3,-2,-1}
\draw[shift={(\x,0)},color=black] (0pt,0pt) -- (0pt,-3pt) node[below]
{$\x$};
\end{tikzpicture}

Cool. Thanks.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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