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The problem you provided contains two vertical angles, \( \angle EFG \) and \( \angle AFC \), and their measures are given as expressions. Since vertical angles are congruent, we can set the expressions equal to each other and solve for \( x \).

Given:
\[
m \angle EFG = x - 2(3 - x)
\]
\[
m \angle AFC = 4 + 2(16 - 2x)
\]

### Step 1: Simplify both expressions.

#### For \( m \angle EFG \):
\[
m \angle EFG = x - 2(3 - x) = x - 6 + 2x = 3x - 6
\]

#### For \( m \angle AFC \):
\[
m \angle AFC = 4 + 2(16 - 2x) = 4 + 32 - 4x = 36 - 4x
\]

### Step 2: Set the two expressions equal since the angles are vertical and congruent:
\[
3x - 6 = 36 - 4x
\]

### Step 3: Solve for \( x \).
First, move all terms involving \( x \) to one side:
\[
3x + 4x = 36 + 6
\]
\[
7x = 42
\]
Now, divide by 7:
\[
x = 6
\]

### Step 4: Verify by substituting \( x = 6 \) back into the original expressions.

For \( m \angle EFG \):
\[
m \angle EFG = 3(6) - 6 = 18 - 6 = 12
\]

For \( m \angle AFC \):
\[
m \angle AFC = 36 - 4(6) = 36 - 24 = 12
\]

Since both angles are equal, the solution \( x = 6 \) is correct.

### Final Answer:
\( x = 6 \).

Would you like further clarification or details on any step?

Here are 5 related questions to explore further:
1. What are vertical angles, and why are they always congruent?
2. How do you simplify expressions with variables and constants?
3. What is the process for solving linear equations?
4. Can two angles with different expressions still be equal?
5. How do we verify solutions by substituting back into the original equations?

**Tip:** Always double-check your work by substituting your solution back into the original equations to ensure accuracy.

Angle EFG and angle AFC are vertical angles. m ∠ EFG = x − 2(3 − x) and m ∠ AFC = 4 + 2(16 − 2x). Find the value of x.

Learn how to solve for x in a vertical angles problem where m ∠ EFG = x − 2(3 − x) and m ∠ AFC = 4 + 2(16 − 2x). This example covers key algebraic concepts such as simplifying expressions and solving linear equations. Suitable for students in Grades 8-10.

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