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solve the inequality lx-3l < (or equal to) 5?
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\(|x-3| \le 5\) With absolute values, yo break it down into 2 situations: 1) if \(x-3 \ge 0\): \(x-3 \le 5\\ x \le 5+3 \\ x \le 8\) 2) if \(x-3 < 0\): \(-(x-3) \le 5 \\ -x+3 \le 5\\ -x \le 2\\ x \ge -2\) So the solution is the union of \(x \le 8\) and \(x \ge -2\) which means: \(-2 \le x \le 8\)
Oh! Thank you for the help! Now it makes more sense.
:)
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