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|3x-2| - 4 <6
pka
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Solve these:
\(\displaystyle \L
\begin{array}{l}
1)\quad 3 = 4 - |2 - x|\quad \Rightarrow \quad |x - 2| = 1 \\
2)\quad |3x - 2| - 4 < 6\quad \Rightarrow \quad |3x - 2| < 10 \\
3)\quad 2 - |3x - 5| < - 4\quad \Rightarrow \quad 6 < |3x - 5| \\
\end{array}\)
\(\displaystyle \L
\begin{array}{l}
1)\quad 3 = 4 - |2 - x|\quad \Rightarrow \quad |x - 2| = 1 \\
2)\quad |3x - 2| - 4 < 6\quad \Rightarrow \quad |3x - 2| < 10 \\
3)\quad 2 - |3x - 5| < - 4\quad \Rightarrow \quad 6 < |3x - 5| \\
\end{array}\)
ButtonNose said:
Add 4 to both sides:
| 3x - 2 | < 10
Now, "absolute value" can be thought of as "distance from 0 on the number line". If something has an absolute value less than 10, then that something must lie between -10 and 10 on the number line. The "something" in this case is 3x - 2. And we know that
-10 < 3x - 2 < 10
Or, you can think in terms of two inequalities:
-10 < 3x - 2 AND 3x - 2 < 10
Can you take it from here?
I hope this helps you.