https://file.aiquickdraw.com/mathbot/file/1726481691995rgrg37e2.png

The image contains two Boolean algebra expressions. Let’s break them down and simplify each.

### Expression 1:
\[
X = \overline{A} \cdot B + A \cdot (\overline{A} + C)
\]
We will simplify this step by step:
1. Apply the Distributive Property to the second term: 
   \[
   A \cdot (\overline{A} + C) = A \cdot \overline{A} + A \cdot C
   \]
   Since \(A \cdot \overline{A} = 0\), we get:
   \[
   A \cdot C
   \]
   So the expression becomes:
   \[
   X = \overline{A} \cdot B + A \cdot C
   \]
   
This is the simplified form for expression 1.

---

### Expression 2:
\[
X = \overline{(A \cdot B \cdot C + D)} \cdot (A \cdot B)
\]
We will simplify this step by step:
1. Use De Morgan's law to break the negation of the sum:
   \[
   \overline{(A \cdot B \cdot C + D)} = \overline{A \cdot B \cdot C} \cdot \overline{D}
   \]
   Using De Morgan’s law on the first term \( \overline{A \cdot B \cdot C} \), we get:
   \[
   \overline{A \cdot B \cdot C} = \overline{A} + \overline{B} + \overline{C}
   \]
   So the expression becomes:
   \[
   X = (\overline{A} + \overline{B} + \overline{C}) \cdot \overline{D} \cdot (A \cdot B)
   \]
2. Distribute \( A \cdot B \) across the terms:
   \[
   X = (A \cdot B \cdot \overline{A}) + (A \cdot B \cdot \overline{B}) + (A

Simplify the following Boolean algebra expressions:
1. X = A̅B + A(A̅ + C)
2. X = (ABC + D)̅AB

Explore step-by-step solutions for simplifying two Boolean algebra expressions. Learn how to apply De Morgan's law and the distributive property for logic simplifications. This problem is suitable for high school and college students.

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