This inequality has a [tex]$\geq$[/tex] symbol:
[tex]\[
\frac{3}{x+2}+4 \geq 3
\][/tex]

But the graph of the solution contains an open circle.

Explain why the solution does not include [tex]$x = -2$[/tex].



Answer :

Certainly! Let's break down the steps to solve the inequality [tex]\(\frac{3}{x+2} + 4 \geq 3\)[/tex] and explain why the solution does not include [tex]\(x = -2\)[/tex].

### Step-by-Step Solution:

1. Subtract 4 from both sides:

[tex]\[\frac{3}{x+2} + 4 - 4 \geq 3 - 4\][/tex]

Simplifying this expression, we get:

[tex]\[\frac{3}{x+2} \geq -1\][/tex]

2. Isolate the term involving [tex]\(x\)[/tex]:

To do this, let's consider the inequality:

[tex]\[ \frac{3}{x+2} \geq -1 \][/tex]

We need to remove the denominator on the left side. We'll multiply both sides of the inequality by [tex]\(x+2\)[/tex], but we must be careful: the sign of [tex]\(x+2\)[/tex] could be positive or negative, which affects our inequality direction.

[tex]\[ \begin{cases} \frac{3}{x+2} \cdot (x+2) \geq -1 \cdot (x+2) & \text{if } x \neq -2 \end{cases} \][/tex]

3. Simplify and solve for [tex]\(x\)[/tex]:

The term [tex]\(\frac{3}{x+2} \cdot (x+2)\)[/tex] simplifies to 3, providing:

[tex]\[ 3 \geq -1(x + 2) \][/tex]

Distribute [tex]\(-1\)[/tex] on the right side:

[tex]\[ 3 \geq -x - 2 \][/tex]

Add 2 to both sides to simplify:

[tex]\[ 3 + 2 \geq -x \][/tex]

[tex]\[ 5 \geq -x \][/tex]

Multiply both sides by [tex]\(-1\)[/tex] (remember to reverse the inequality sign):

[tex]\[ -5 \leq x \][/tex]

Or,

[tex]\[ x \geq -5 \][/tex]

4. Consider [tex]\(x = -2\)[/tex]:

Throughout these steps, an important factor is that [tex]\(x \neq -2\)[/tex] because if [tex]\(x = -2\)[/tex], the denominator in the original inequality [tex]\(\frac{3}{x+2} + 4 \geq 3\)[/tex] would become zero, which would make the fraction undefined.

[tex]\[ \frac{3}{x + 2} \quad \text{is undefined when } x = -2 \][/tex]

### Conclusion:

The solution to the inequality [tex]\(\frac{3}{x+2} + 4 \geq 3\)[/tex] is [tex]\(x \geq -5\)[/tex], but not including [tex]\(x = -2\)[/tex], because at [tex]\(x = -2\)[/tex] the term [tex]\(\frac{3}{x+2}\)[/tex] is undefined (it involves division by zero).

Thus, the range of [tex]\(x\)[/tex] is:
[tex]\[ x \in [-5, \infty) \quad \text{excluding} \quad x = -2 \][/tex]

This exclusion of [tex]\(x = -2\)[/tex] is why there is an open circle at [tex]\(x = -2\)[/tex] on the graph of the solution. An open circle represents a value that is not included in the solution set.