To find the integers $n$ that satisfy both inequalities, I need to find the intersection of the solution sets of each inequality. I can do this by solving each inequality separately and then finding the common values of $n$. To solve the first inequality, $4n + 3 < 25$, I can subtract 3 from both sides and then divide by 4, getting $n < 5.5$. To solve the second inequality, $-7n + 5 < 24$, I can subtract 5 from both sides and then divide by -7, reversing the inequality sign, getting $n > -2.71$. So the solution set of the first inequality is $(-\infty, 5.5)$ and the solution set of the second inequality is $(-2.71, \infty)$. The intersection of these sets is $(-2.71, 5.5)$, which means that any value of $n$ between -2.71 and 5.5, exclusive, will satisfy both inequalities. However, the question asks for the integers $n$ that satisfy both inequalities, not the real numbers $n$. So I need to find the integers that are in the interval $(-2.71, 5.5)$. I can see that the smallest integer in this interval is -2, and the largest integer is 5. So the possible values of $n$ are -2, -1, 0, 1, 2, 3, 4, and 5. There are 8 such values. 
How many integers $n$ satisfy both of the inequalities $4n + 3 < 25$ and $-7n + 5 < 24$?