5 /leq 4x + 1 < 25

Answer 1

To solve the compound inequality \( \frac{5}{4} \leq 4x + 1 < 25 \), we can break it down into two parts and solve each part step by step.

Part 1: Solve \( \frac{5}{4} \leq 4x + 1 \)

1. Subtract 1 from both sides: \[ \frac{5}{4} - 1 \leq 4x \]

2. Convert 1 into a fraction with a common denominator: \[ \frac{5}{4} - \frac{4}{4} = \frac{1}{4} \] So we have: \[ \frac{1}{4} \leq 4x \]

3. Divide both sides by 4: \[ \frac{1}{16} \leq x \]

Part 2: Solve \( 4x + 1 < 25 \)

1. Subtract 1 from both sides: \[ 4x < 25 - 1 \] \[ 4x < 24 \]

2. Divide both sides by 4: \[ x < 6 \]

Combine the results

From the parts we solved, we have: \[ \frac{1}{16} \leq x < 6 \]

Final Answer

The solution to the compound inequality is: \[ \frac{1}{16} \leq x < 6 \]

In interval notation, this is: \[ \left[ \frac{1}{16}, 6 \right) \]