__Solution to example 1__

The above inequality is solved by writing a double inequality equivalent to the given inequality but without absolute value

\(- 3 \lt x + 2 \lt 3 \)

Solve the double inequality to obtain

\(- 5 \lt x \lt 1 \)

The above solution set is written in interval form as follows

\((-5 , 1)\)

__Example 2:__ Solve the inequality.

\(|- 2 x - 4| \gt 9\)
__Solution to example 2__

Solving the above inequality is equivalent to solving

\(-2 x - 4 \gt 9\) or \(-2 x - 4 \lt - 9\)

Which gives

\( x \lt -13 / 2\) or \( x \gt 5 / 2 \)

The above solution set is written in interval form as follows

\( (-\infty , -13 / 2) \cup (5 / 2 , + \infty) \)

__Example 3:__ Solve the inequality.

\( x + 2 \lt |x^2 - 4| \)
__Solution to example 3__

__Condition 1__ - For \( x^2 - 4 \gt = 0 \), or \( x\) in the interval \( (-\infty , -2] \cup [2 , +\infty)\), we can write

\( |x^2 - 4| = x^2 - 4 \)

Substitute the expression with the absolute value in the given inequality and solve

\( x + 2 \lt x^2 - 4\)

\( x^2 - x - 6 \gt 0\)

The solution set to above inequality is given by the interval

\( (-\infty , -2) \cup (3 , +\infty)\)

The intersection of intervals \((-\infty , -2] \cup [2 , +\infty)\) and \((-\infty , -2) \cup (3 , +\infty)\) gives the solution set

\((-\infty , -2) \cup (3 , +\infty)\)

__Condition 2__ - For \(x^2 - 4 \lt 0\), or \(x\) in the interval \((-2 , 2)\), we can write

\( |x^2 - 4| = -(x^2 - 4) \)

Substitute the expression with the absolute value in the given inequality and solve

\(x + 2 \lt -(x^2 - 4)\)

\(x^2 + x - 2 \lt 0\)

The solution set to above inequatlity is given by the intersection of the intervals \((-2 , 1)\) and \((-2 , 2)\) which gives

(-2 , 1)

Conclusion: The solution set to the given inequality is
\((-\infty, -2) \cup (-2 , 1) \cup (3 , +\infty)\)

Check the above answer to the inequality graphically.(see graph below).

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