Question

solve algebraically.

1. |2x - 5| ≤ 9
2. |4x + 1| > 7

Explanation:

Problem 17: Solve \(|2x - 5| \leq 9\)

Step 1: Recall the absolute value inequality property

For \(|a| \leq b\) (where \(b\geq0\)), it is equivalent to \(-b \leq a \leq b\). So for \(|2x - 5| \leq 9\), we have \(-9 \leq 2x - 5 \leq 9\).

Step 2: Add 5 to all parts of the inequality

Adding 5 to each part: \(-9 + 5 \leq 2x - 5 + 5 \leq 9 + 5\), which simplifies to \(-4 \leq 2x \leq 14\).

Step 3: Divide all parts by 2

Dividing each part by 2: \(\frac{-4}{2} \leq \frac{2x}{2} \leq \frac{14}{2}\), so \(-2 \leq x \leq 7\).

Step 1: Recall the absolute value inequality property

For \(|a| > b\) (where \(b\geq0\)), it is equivalent to \(a > b\) or \(a < -b\). So for \(|4x + 1| > 7\), we have two cases: \(4x + 1 > 7\) or \(4x + 1 < -7\).

Step 2: Solve the first inequality \(4x + 1 > 7\)

Subtract 1 from both sides: \(4x + 1 - 1 > 7 - 1\), which gives \(4x > 6\). Then divide by 4: \(x > \frac{6}{4}=\frac{3}{2}\).

Step 3: Solve the second inequality \(4x + 1 < -7\)

Subtract 1 from both sides: \(4x + 1 - 1 < -7 - 1\), which gives \(4x < -8\). Then divide by 4: \(x < \frac{-8}{4}=-2\).

Answer:

\(-2 \leq x \leq 7\)

Problem 18: Solve \(|4x + 1| > 7\)