Find all the values that make the inequality x+4/x-3>or equal to x-5/x+2 true and write your answer in interval notation or set notation.

Answer 1

To solve the inequality \(\frac{x+4}{x-3} \geq \frac{x-5}{x+2}\), follow these steps:

1. **Combine the fractions:**
We start by moving to a common denominator.

\[
\frac{x+4}{x-3} - \frac{x-5}{x+2} \geq 0
\]

2. **Combine the rational expressions:**
\[
\frac{(x+4)(x+2) - (x-5)(x-3)}{(x-3)(x+2)} \geq 0
\]

3. **Expand and simplify the numerator:**

Expand \((x+4)(x+2)\):
\[
x^2 + 2x + 4x + 8 = x^2 + 6x + 8
\]

Expand \((x-5)(x-3)\):
\[
x^2 - 3x - 5x + 15 = x^2 - 8x + 15
\]

Simplify the numerator:
\[
x^2 + 6x + 8 - (x^2 - 8x + 15) = x^2 + 6x + 8 - x^2 + 8x - 15 = 14x - 7
\]

So the inequality becomes:
\[
\frac{14x - 7}{(x-3)(x+2)} \geq 0
\]

4. **Factor the numerator:**
\[
14x - 7 = 7(2x - 1)
\]

The inequality can thus be written as:
\[
\frac{7(2x - 1)}{(x-3)(x+2)} \geq 0
\]

5. **Find critical points where the expression is zero or undefined:**

- The numerator \(7(2x - 1) = 0\) when \(2x - 1 = 0\), hence \(x = \frac{1}{2}\).
- The denominator \((x-3)(x+2) = 0\) when \(x = 3\) or \(x = -2\).

6. **Determine the intervals:**

Consider the points \(x = -2\), \(x = \frac{1}{2}\), and \(x = 3\). These divide the number line into the following intervals:
- \((-\infty, -2)\),
- \((-2, \frac{1}{2})\),
- \((\frac{1}{2}, 3)\),
- \((3, \infty)\).

7. **Test each interval:**

Check the sign of \(\frac{7(2x - 1)}{(x-3)(x+2)}\) in each interval:

- For \(x \in (-\infty, -2)\): pick \(x = -3\).

\[
\frac{7(2(-3) - 1)}{((-3)-3)((-3)+2)} = \frac{7(-7)}{-6} = \frac{-49}{-6} > 0
\]

- For \(x \in (-2, \frac{1}{2})\): pick \(x = 0\).

\[
\frac{7(2(0) - 1)}{((0)-3)((0)+2)} = \frac{7(-1)}{-6} = \frac{-7}{-6} > 0
\]

- For \(x \in (\frac{1}{2}, 3)\): pick \(x = 1\).

\[
\frac{7(2(1) - 1)}{(1-3)(1+2)} = \frac{7(1)}{-6} = \frac{7}{-6} < 0
\]

- For \(x \in (3, \infty)\): pick \(x = 4\).

\[
\frac{7(2(4)-1)}{(4-3)(4+2)} = \frac{7(7)}{6} > 0
\]

8. **Include critical points and endpoints:**

- \(x = \frac{1}{2}\) satisfies the inequality because \(\frac{7(2(\frac{1}{2}) - 1)}{((\frac{1}{2})-3)((\frac{1}{2})+2)} = \frac{7(0)}{...} = 0\).
- \(x = -2\) or \(x = 3\) make the expression undefined and should not be included.

So, the solution in interval notation is:

\[
(-\infty, -2) \cup \left( -2, \frac{1}{2} \right] \cup (3, \infty)
\]