Solve the compound inequality -2 < (2x - 3)/(x + 1) <= 1.

Step-by-step solution

1. 1

   Do not multiply through by $x+1$. Its sign is unknown, so multiplying would require case-splitting and risks flipping the inequality. Instead split the chain into two separate inequalities and note the domain restriction $x\ne-1$ up front.

2. 2

   Solve the left half by moving everything to one side.

   $\frac{2x-3}{x+1}+2>0\;\Longrightarrow\;\frac{2x-3+2(x+1)}{x+1}>0\;\Longrightarrow\;\frac{4x-1}{x+1}>0$

   Critical points $x=\tfrac14$ and $x=-1$. Testing $x=-2$ gives $\tfrac{-9}{-1}>0$, $x=0$ gives $-1<0$, $x=1$ gives $\tfrac32>0$, so the left half holds on $(-\infty,-1)\cup\left(\tfrac14,\infty\right)$.

3. 3

   Solve the right half the same way.

   $\frac{2x-3}{x+1}-1\le0\;\Longrightarrow\;\frac{2x-3-(x+1)}{x+1}\le0\;\Longrightarrow\;\frac{x-4}{x+1}\le0$

   Critical points $x=4$ (allowed, since equality is permitted) and $x=-1$ (never allowed). Testing $x=-2$ gives $6>0$, $x=0$ gives $-4<0$, $x=5$ gives $\tfrac16>0$, so this half holds on $(-1,4]$.

4. 4

   Intersect the two solution sets. A number must satisfy both halves:

   $\left[(-\infty,-1)\cup\left(\tfrac14,\infty\right)\right]\cap(-1,4]=\left(\tfrac14,\,4\right]$

   The branch $(-\infty,-1)$ is wiped out because the right half excludes everything below $-1$.

5. 5

   Test the boundaries numerically. At $x=0.26$ the quotient is $-1.968$, which satisfies $-2<\cdot\le1$; at $x=0.2$ it is $-2.167$, which fails on the left. At $x=4$ the quotient is exactly $1$ (included); at $x=4.1$ it is $1.0196$, which fails on the right. The endpoints behave exactly as claimed.

6. 6

   Write the final interval.

   $x\in\left(\frac14,\,4\right]$