How many pairs of integers $(b,c)$ satisfy the equation

\[\frac{b + 7}{b + 4} = \frac{c}{9}?\]

On sait que x+1/x>2

To solve this equation, we can start by cross-multiplying:

\[(b + 7) \cdot 9 = (b + 4) \cdot c.\]

Expanding both sides of the equation gives:

\[9b + 63 = cb + 4c.\]

Rearranging terms, we get:

\[cb - 9b = 4c - 63.\]

Factoring out a $b$ on the left side of the equation gives:

\[b(c - 9) = 4c - 63.\]

Now, we can solve for $b$:

\[b = \frac{4c - 63}{c - 9}.\]

For each integer value of $c$, we can substitute it into this equation to find the corresponding $b$. However, we must also check the condition that $b + 4$ is not equal to $0$, as this would result in a division by zero, which is undefined.

Therefore, to find the number of pairs $(b,c)$ that satisfy the given equation, we need to find the values of $c$ that make $b$ an integer, considering the restriction that $b + 4$ is not equal to $0$.

To solve the equation, we can start by cross multiplying:

\[9(b+7) = c(b+4).\]
Expanding the equation, we get:
\[9b + 63 = cb + 4c.\]
Rearranging the terms, we have:
\[cb - 9b = 4c - 63.\]
Factoring out $b$ on the left-hand side, we have:
\[b(c-9) = 4c - 63.\]
Now, let's consider the cases when $c = 9$ and $c \neq 9$ separately:

Case 1: $c = 9$
When $c = 9,$ the equation becomes:
\[b(9-9) = 4\cdot9 - 63.\]
Simplifying, we have:
\[0 = 36 - 63.\]
However, this equation is not true. Therefore, there are no solutions in this case.

Case 2: $c \neq 9$
When $c \neq 9,$ we can divide both sides of the equation by $(c-9):$
\[b = \frac{4c - 63}{c-9}.\]
Since we're looking for integer solutions, we need to determine when $\frac{4c - 63}{c-9}$ is an integer.
Notice that if $\frac{4c - 63}{c-9}$ is an integer, then $4c - 63$ is divisible by $c-9.$
Using the Remainder Theorem, we can divide the polynomial $4c - 63$ by $c-9$ and set the remainder equal to zero.

To perform the polynomial division, let's write $4c - 63$ as $(c-9)q(r),$ where $q$ is the quotient and $r$ is the remainder. Since the divisor is linear, the remainder has degree 0, or in other words, $r$ must be a constant. Thus, we will be left with:
\[4c - 63 = (c-9)q.\]
Expanding the equation, we get:
\[4c - 63 = cq - 9q.\]
Rearranging the terms, we have:
\[q(c-9) = 4c - 63.\]
Just as before, the left-hand side must be divisible by $c-9.$ Therefore, we must have:
\[4c - 63 = 0.\]
Solving this equation, we find that $c = \frac{63}{4},$ which is not an integer. Therefore, there are no solutions in this case.

In conclusion, there are $\boxed{0}$ pairs of integers $(b,c)$ that satisfy the equation.