Of $ 28$ students taking at least one subject the number taking Mathematics and English only equals the number taking Mathematics only.  No student takes English only or History only, and six students take Mathematics and History, but not English.  The number taking English and History only is five times the number taking all three subjects.  If the number taking all three subjects is even and non-zero, the number taking English and Mathematics only is:

$ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$
1. Define the variables for the number of students taking each combination of subjects:
   - Let \( x \) be the number of students taking Mathematics and English only.
   - Let \( y \) be the number of students taking all three subjects (Mathematics, English, and History).
   - Let \( z \) be the number of students taking English and History only.

2. Given conditions:
   - The number of students taking Mathematics and English only equals the number taking Mathematics only: \( x \).
   - No student takes English only or History only: \( 0 \).
   - Six students take Mathematics and History, but not English: \( 6 \).
   - The number taking English and History only is five times the number taking all three subjects: \( z = 5y \).
   - The total number of students is 28.

3. Using the principle of inclusion-exclusion for three sets, we have:
   \[
   28 = (M) + (E) + (H) - (ME) - (MH) - (EH) + (MEH)
   \]
   where:
   - \( (M) \) is the number of students taking Mathematics only: \( x \).
   - \( (E) \) is the number of students taking English only: \( 0 \).
   - \( (H) \) is the number of students taking History only: \( 0 \).
   - \( (ME) \) is the number of students taking Mathematics and English only: \( x \).
   - \( (MH) \) is the number of students taking Mathematics and History only: \( 6 \).
   - \( (EH) \) is the number of students taking English and History only: \( 5y \).
   - \( (MEH) \) is the number of students taking all three subjects: \( y \).

4. Substitute the values into the inclusion-exclusion formula:
   \[
   28 = x + 0 + 0 - x - 6 - 5y + y
   \]
   Simplify the equation:
   \[
   28 = x - x - 6 - 5y + y
   \]
   \[
   28 = -6 - 4y
   \]
   \[
   28 + 6 = -4y
   \]
   \[
   34 = -4y
   \]
   \[
   y = -\frac{34}{4}
   \]
   \[
   y = -8.5
   \]

5. Since \( y \) must be an even and non-zero positive integer, we need to re-evaluate the steps. Let's correct the equation:
   \[
   28 = 2x + 6 + 6y
   \]
   \[
   28 = 2x + 6 + 6y
   \]
   \[
   22 = 2x + 6y
   \]
   \[
   11 = x + 3y
   \]

6. Since \( y \) is an even and non-zero positive integer, the possible values for \( y \) are 2, 4, 6, etc. Let's test \( y = 2 \):
   \[
   11 = x + 3(2)
   \]
   \[
   11 = x + 6
   \]
   \[
   x = 5
   \]

7. Therefore, the number of students taking Mathematics and English only is \( x = 5 \).

The final answer is \( \boxed{5} \).