To solve the compound inequality \(4 - 2x > 7\) and \(2 - 3(x + 1) < 5\), we will solve each inequality separately.
First Inequality:
\[ 4 - 2x > 7 \]
Subtract 4 from both sides:
\[ -2x > 7 - 4 \]
This simplifies to:
\[ -2x > 3 \]
Now, divide both sides by -2. Remember to flip the inequality sign when dividing by a negative number:
\[ x < -\frac{3}{2} \]
Second Inequality:
\[ 2 - 3(x + 1) < 5 \]
First, distribute the -3:
\[ 2 - 3x - 3 < 5 \]
Combine like terms:
\[ -3x - 1 < 5 \]
Add 1 to both sides:
\[ -3x < 5 + 1 \]
This simplifies to:
\[ -3x < 6 \]
Now, divide both sides by -3, remembering to flip the inequality sign:
\[ x > -2 \]
Combining Results:
We have the two inequalities:
- \(x < -\frac{3}{2}\)
- \(x > -2\)
The solution must satisfy both conditions. So we can express the solution in interval notation:
\[ -2 < x < -\frac{3}{2} \]
This is the solution to the compound inequality. In interval notation, it is:
\[ (-2, -\frac{3}{2}) \]