The problem asks us to solve two inequalities for $n$. The first inequality is $\frac{n-1}{n+1} < 5$ and the second inequality is $1 < \frac{3n+10}{n+7} < 2$.

1. Problem Description

The problem asks us to solve two inequalities for $n$. The first inequality is $\frac{n-1}{n+1} < 5$ and the second inequality is $1 < \frac{3n+10}{n+7} < 2$.

2. Solution Steps

a) Solving the inequality $\frac{n-1}{n+1} < 5$:

Subtract 5 from both sides:

$\frac{n-1}{n+1} - 5 < 0$

Find a common denominator:

$\frac{n-1 - 5(n+1)}{n+1} < 0$

Simplify the numerator:

$\frac{n-1 - 5n - 5}{n+1} < 0$

$\frac{-4n - 6}{n+1} < 0$

Multiply both sides by -1, and reverse the inequality sign:

$\frac{4n+6}{n+1} > 0$

To solve this inequality, we find the critical points: $4n+6 = 0 \Rightarrow n = -\frac{6}{4} = -\frac{3}{2}$ and $n+1 = 0 \Rightarrow n = -1$.

Now, we consider the intervals determined by these critical points: $(-\infty, -\frac{3}{2})$, $(-\frac{3}{2}, -1)$, and $(-1, \infty)$.

For $n < -\frac{3}{2}$, let $n = -2$. Then $\frac{4(-2)+6}{-2+1} = \frac{-8+6}{-1} = \frac{-2}{-1} = 2 > 0$. So the inequality holds.

For $-\frac{3}{2} < n < -1$, let $n = -1.25$. Then $\frac{4(-1.25)+6}{-1.25+1} = \frac{-5+6}{-0.25} = \frac{1}{-0.25} = -4 < 0$. So the inequality does not hold.

For $n > -1$, let $n = 0$. Then $\frac{4(0)+6}{0+1} = \frac{6}{1} = 6 > 0$. So the inequality holds.

Therefore, the solution to the first inequality is $n < -\frac{3}{2}$ or $n > -1$.

b) Solving the inequality $1 < \frac{3n+10}{n+7} < 2$:

We have two inequalities:

$1 < \frac{3n+10}{n+7}$ and $\frac{3n+10}{n+7} < 2$.

For $1 < \frac{3n+10}{n+7}$:

$1 - \frac{3n+10}{n+7} < 0$

$\frac{n+7 - (3n+10)}{n+7} < 0$

$\frac{n+7 - 3n - 10}{n+7} < 0$

$\frac{-2n - 3}{n+7} < 0$

$\frac{2n+3}{n+7} > 0$

Critical points: $2n+3 = 0 \Rightarrow n = -\frac{3}{2}$ and $n+7 = 0 \Rightarrow n = -7$.

Intervals: $(-\infty, -7)$, $(-7, -\frac{3}{2})$, and $(-\frac{3}{2}, \infty)$.

For $n < -7$, let $n = -8$. $\frac{2(-8)+3}{-8+7} = \frac{-16+3}{-1} = \frac{-13}{-1} = 13 > 0$.

For $-7 < n < -\frac{3}{2}$, let $n = -2$. $\frac{2(-2)+3}{-2+7} = \frac{-4+3}{5} = \frac{-1}{5} < 0$.

For $n > -\frac{3}{2}$, let $n = 0$. $\frac{2(0)+3}{0+7} = \frac{3}{7} > 0$.

Thus, the solution to $1 < \frac{3n+10}{n+7}$ is $n < -7$ or $n > -\frac{3}{2}$.

For $\frac{3n+10}{n+7} < 2$:

$\frac{3n+10}{n+7} - 2 < 0$

$\frac{3n+10 - 2(n+7)}{n+7} < 0$

$\frac{3n+10 - 2n - 14}{n+7} < 0$

$\frac{n-4}{n+7} < 0$

Critical points: $n-4 = 0 \Rightarrow n = 4$ and $n+7 = 0 \Rightarrow n = -7$.

Intervals: $(-\infty, -7)$, $(-7, 4)$, and $(4, \infty)$.

For $n < -7$, let $n = -8$. $\frac{-8-4}{-8+7} = \frac{-12}{-1} = 12 > 0$.

For $-7 < n < 4$, let $n = 0$. $\frac{0-4}{0+7} = \frac{-4}{7} < 0$.

For $n > 4$, let $n = 5$. $\frac{5-4}{5+7} = \frac{1}{12} > 0$.

Thus, the solution to $\frac{3n+10}{n+7} < 2$ is $-7 < n < 4$.

Now, we need to find the intersection of $n < -7$ or $n > -\frac{3}{2}$ and $-7 < n < 4$.

Since $n < -7$ and $-7 < n < 4$ cannot happen at the same time, we consider $n > -\frac{3}{2}$ and $-7 < n < 4$.

The intersection is $-\frac{3}{2} < n < 4$.

3. Final Answer

a) $n < -\frac{3}{2}$ or $n > -1$

b) $-\frac{3}{2} < n < 4$