Inequalities and Absolute Value Practice Questions

When solving an absolute value equation, it helps to first isolate the absolute value expression:<br><br>\( 3|3 - x| = 7 \)<br>\( |3 - x| = \frac{7}{3} \)<br><br>When removing the absolute value bars, we need to keep in mind that the expression inside the absolute value bars \( (3 - x) \) could be positive or negative. Let's consider both possibilities:<br><br>When \( (3 - x) \) is positive:<br>\( (3 - x) = \frac{7}{3} \)<br>\( 3 - \frac{7}{3} = x \)<br>\( \frac{9}{3} - \frac{7}{3} = x \)<br>\( x = \frac{2}{3} \)<br><br>When \( (3 - x) \) is negative:<br>\( -(3 - x) = \frac{7}{3} \)<br>\( x - 3 = \frac{7}{3} \)<br>\( x = \frac{7}{3} + 3 \)<br>\( x = \frac{7}{3} + \frac{9}{3} \)<br>\( x = \frac{16}{3} \)<br>So, the two possible values for \( x \) are \( \frac{2}{3} \) and \( \frac{16}{3} \). The product of these values is \( \frac{32}{9} \).<br>The correct answer is E.

The possible values for \( a \) and \( b \) are \( \pm \frac{1}{3} \) and \( \pm \frac{2}{3} \) respectively.<br>We can list the possible sums:<br><br>$$\begin{array}{|c|c|c|} \hline a & b & a+b \\ \hline \frac{1}{3} & \frac{2}{3} & 1 \\ \hline \frac{1}{3} & -\frac{2}{3} & -\frac{1}{3} \\ \hline -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ \hline -\frac{1}{3} & -\frac{2}{3} & -1 \\ \hline \end{array}$$<br><br>Comparing with the options, \( \frac{2}{3} \) is NOT a possible sum.<br>The correct answer is D.

Simplify the equation: \( \sqrt{(x + 4)^2} = |x + 4| = 3 \).<br><br>This yields two cases:<br>\( x + 4 = 3 \implies x = -1 \)<br>\( x + 4 = -3 \implies x = -7 \)<br><br>The question asks for the value of \( x - 4 \), not \( x \) itself. Calculate \( x - 4 \) for both solutions:<br>If \( x = -1 \): \( x - 4 = -1 - 4 = -5 \) (not among the answer choices).<br>If \( x = -7 \): \( x - 4 = -7 - 4 = -11 \).<br><br>Since \( -11 \) is the only value that appears among the answer choices, the correct answer is A.