To determine the valid expression for the possible values of \( n \) for the sides of a triangle, we can use the triangle inequality theorem. The triangle inequality theorem states that for any triangle with side lengths \(a\), \(b\), and \(c\):
1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)
In this problem, the sides of the triangle are given as:
[tex]\[
a = 2x + 2, \quad b = x + 3, \quad \text{and} \quad c = n
\][/tex]
Applying the triangle inequality theorem, we get the following three inequalities:
1. \( (2x + 2) + (x + 3) > n \)
2. \( (2x + 2) + n > x + 3 \)
3. \( (x + 3) + n > 2x + 2 \)
We will solve each inequality step-by-step:
### Inequality 1: \( (2x + 2) + (x + 3) > n \)
[tex]\[
2x + 2 + x + 3 > n
\][/tex]
[tex]\[
3x + 5 > n
\][/tex]
or equivalently,
[tex]\[
n < 3x + 5
\][/tex]
### Inequality 2: \( (2x + 2) + n > x + 3 \)
[tex]\[
2x + 2 + n > x + 3
\][/tex]
[tex]\[
2x + 2 + n - x > 3
\][/tex]
[tex]\[
x + 2 + n > 3
\][/tex]
[tex]\[
n > 3 - x - 2
\][/tex]
[tex]\[
n > x - 1
\][/tex]
### Inequality 3: \( (x + 3) + n > 2x + 2 \)
[tex]\[
x + 3 + n > 2x + 2
\][/tex]
[tex]\[
n > 2x + 2 - x - 3
\][/tex]
[tex]\[
n > x - 1
\][/tex]
Even though we derived the same inequality for the second part, it reconfirms our previous finding.
Thus, combining both inequalities from above, we get:
[tex]\[
x - 1 < n < 3x + 5
\][/tex]
This means that \( n \) must be greater than \( x - 1 \) and less than \( 3x + 5 \) to satisfy all three conditions of the triangle inequality.
The correct solution is:
[tex]\[
x - 1 < n < 3x + 5
\][/tex]
Thus, the correct answer is:
[tex]\[
\boxed{x - 1 < n < 3x + 5}
\][/tex]
