Example Question - solving absolute value
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Solving Absolute Value Inequalities with Two Cases
To solve the inequality \(7 \leq |7y - 9|\), we must consider two cases because the absolute value function \(|x|\) is defined as \(x\) when \(x \geq 0\) and \(-x\) when \(x < 0\). **Case 1**: \(7y - 9 \geq 0\) If \(7y - 9\) is non-negative, the inequality is simply: \(7 \leq 7y - 9\) Now, we solve for \(y\): \(7 + 9 \leq 7y\) \(16 \leq 7y\) Divide both sides by 7: \(y \geq \frac{16}{7}\) **Case 2**: \(7y - 9 < 0\) If \(7y - 9\) is negative, then \(|7y - 9| = -(7y - 9)\), so the inequality becomes: \(7 \leq -(7y - 9)\) Now, we solve for \(y\): \(7 \leq -7y + 9\) Rearrange: \(7y \leq 9 - 7\) \(7y \leq 2\) Divide both sides by 7: \(y \leq \frac{2}{7}\) Combining the results from both cases gives us the complete solution set for the inequality: \(\frac{2}{7} \geq y \geq \frac{16}{7}\) However, if you look carefully at the inequalities derived from both cases, you'll notice that there's actually no overlap between the two solution sets, since it's not possible for a number \(y\) to be simultaneously greater than or equal to approximately \(2.29\) and less than or equal to approximately \(0.29\). This suggests we need to correct the combined inequality to reflect the actual solution. The correct combined solution for the inequality considering both cases would be: \(y \leq \frac{2}{7}\) or \(y \geq \frac{16}{7}\) Thus, the solution set for the inequality \(7 \leq |7y - 9|\) is that \(y\) can be in the interval \((-\infty, \frac{2}{7}]\) or \([\frac{16}{7}, +\infty)\).
Absolute Value Inequality Solution
The inequality is \( 3|d| + 5 < 47 \). First, we'll isolate the absolute value expression by subtracting 5 from both sides of the inequality: \( 3|d| + 5 - 5 < 47 - 5 \) \( 3|d| < 42 \) Now, divide both sides by 3 to solve for the absolute value of d: \( \frac{3|d|}{3} < \frac{42}{3} \) \( |d| < 14 \) Since we have an absolute value inequality, we know that \( |d| < 14 \) means that d is less than 14 and greater than -14. Therefore, the solution set for d is: \( -14 < d < 14 \)
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