Question

Solve the compound inequality.
4 - 2x > 7 and 2 - 3(x + 1) < 5

Answer 1

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:

1. \(x < -\frac{3}{2}\)
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:

\[
(-2, -\frac{3}{2})
\]