We are given the following conditions in the problem:

• $\overline{AB} \parallel \overline{CD}$,
• $EF = FG$,
• $\angle AEG = x^\circ$,
• $\angle BEF = 100^\circ + x^\circ$.

Our goal is to find the value of $x$.

Step 1: Analyze the angles

From the diagram, we know that lines $\overline{AB}$ and $\overline{CD}$ are parallel, and point $E$ lies on line $\overline{AB}$. Point $F$ lies on line $\overline{CD}$, and points $F$ and $G$ lie along the transversal $EG$.

Thus, since $\overline{AB} \parallel \overline{CD}$, angles that are formed by the transversal are related. Specifically, we observe the angles at $E$ and $F$.

Step 2: Use the given angle information

We are told that:

• $\angle AEG = x^\circ$,
• $\angle BEF = 100^\circ + x^\circ$.

Step 3: Consider the external angles

Since $EF = FG$, triangle $EFG$ is isosceles. This means that the base angles, $\angle BEF$ and $\angle GEF$, are equal. Thus, we have: $\angle GEF = 100^\circ + x^\circ.$

Step 4: Sum of angles in a triangle

Now, consider triangle $AEG$. The sum of the angles in any triangle must be $180^\circ$. Therefore, we can write the equation for triangle $AEG$ as: $\angle AEG + \angle GEF + \angle EAG = 180^\circ.$ Substitute the known values: $x^\circ + (100^\circ + x^\circ) + \angle EAG = 180^\circ.$ Simplifying: $2x^\circ + 100^\circ + \angle EAG = 180^\circ.$ This simplifies to: $2x + \angle EAG = 80^\circ.$

Step 5: Solve for $x$

Since $\angle EAG$ must be a positive angle, we now solve this equation to find the value of $x$.