To solve the system of equations given:

1. \( ad - cb = 11 \) (Equation 1)
2. \( ad \times cb = 2622 \) (Equation 2)

Let:
- \( x = ad \)
- \( y = cb \)

Now we can rewrite the equations as:
1. \( x - y = 11 \) (Equation 1)
2. \( xy = 2622 \) (Equation 2)

From Equation 1, solve for \( x \):
\[ x = y + 11 \]

Substituting this into Equation 2:
\[ (y + 11)y = 2622 \]
\[ y^2 + 11y - 2622 = 0 \]

Now we can use the quadratic formula to solve for \( y \):
\[ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
where \( a = 1 \), \( b = 11 \), and \( c = -2622 \).

Calculating the discriminant:
\[ b^2 - 4ac = 11^2 - 4 \times 1 \times (-2622) = 121 + 10488 = 10509 \]

Now, calculating \( y \):
\[ y = \frac{{-11 \pm \sqrt{10509}}}{2} \]

Calculating \( \sqrt{10509} \):
\[ \sqrt{10509} \approx 102.5 \]

Now substituting back in:
\[ y \approx \frac{{-11 \pm 102.5}}{2} \]

Calculating the two possible values:
1. \( y \approx \frac{2} \approx 45.75 \) (which we will discard because \( y \) should be an integer)
2. \( y \approx \frac{{-113.5}}{2} \) (not feasible)

Since \( y \) must be an integer, let's find the integer factors of 2622:
The factor pairs of 2622 are \( (1, 2622), (2, 1311), (6, 437), (11, 242), (22, 119), (33, 79) \).

We can find suitable combinations that satisfy:
1. \( x = y + 11 \)
2. \( xy = 2622 \)

Using integer pairs:
- For \( y = 42 \) (from pairs), \( x = y + 11 = 53 \).
- Check if \( 53 \times 42 = 2622 \).

So, \( ad = 53 \) and \( cb = 42 \).

Thus, the solutions are:
- \( ad = 53 \)
- \( cb = 42 \)

These satisfy both equations!