To find a value that is a solution to two of the equations, we need to solve each equation for its variable.
1. **For the first equation:**
\( \frac{p}{2} + 4 = 10 \)
Subtract 4 from both sides:
\( \frac{p}{2} = 6 \)
Multiply both sides by 2:
\( p = 12 \)
2. **For the second equation:**
\( 2 = 10 - 6h \)
Subtract 10 from both sides:
\( -8 = -6h \)
Divide by -6:
\( h = \frac{4}{3} \)
3. **For the third equation:**
\( 3y - 16 = 20 \)
Add 16 to both sides:
\( 3y = 36 \)
Divide by 3:
\( y = 12 \)
Now we can list the solutions we found:
- The solution for the first equation is \( p = 12 \).
- The solution for the second equation is \( h = \frac{4}{3} \).
- The solution for the third equation is \( y = 12 \).
The common value that is a solution to two of the equations is **12**, as it is a solution to both the first and third equations.
Thus, the solution set is:
\[
\{12\}
\]