A triangle has side lengths measuring [tex]$2 x+2 ft$[/tex], [tex]$x+3 ft$[/tex], and [tex]$n ft$[/tex]. Which expression represents the possible values of [tex]$n$[/tex], in feet? Express your answer in simplest terms.
[tex]$x-1\ \textless \ n\ \textless \ 3 x+5$[/tex]

Apply the triangle inequality theorem to the given side lengths: 2 x + 2 , x + 3 , and n .
Establish three inequalities based on the theorem: n"> ( 2 x + 2 ) + ( x + 3 ) > n , (x+3)"> ( 2 x + 2 ) + n > ( x + 3 ) , and (2x+2)"> ( x + 3 ) + n > ( 2 x + 2 ) .
Simplify the inequalities to find the range of possible values for n : n"> 3 x + 5 > n , -x+1"> n > − x + 1 , and x-1"> n > x − 1 .
Combine these inequalities with the given condition x − 1 < n < 3 x + 5 to determine the final range: ∣ x − 1∣ < n < 3 x + 5 .
The expression representing the possible values of n is ∣ x − 1∣ < n < 3 x + 5 ​ .

Explanation

Problem Analysis We are given a triangle with side lengths 2 x + 2 , x + 3 , and n . We are also given that x − 1 < n < 3 x + 5 . We need to find the possible values of n .

Applying the Triangle Inequality The triangle inequality states that the sum of any two sides of a triangle must be greater than the third side. Therefore, we have the following inequalities:

n"> ( 2 x + 2 ) + ( x + 3 ) > n

(x+3)"> ( 2 x + 2 ) + n > ( x + 3 )

(2x+2)"> ( x + 3 ) + n > ( 2 x + 2 )

Simplifying the Inequalities Simplifying the inequalities:

n"> 3 x + 5 > n

-x + 1"> n > − x + 1

x - 1"> n > x − 1

Combining the Inequalities and Given Conditions Combining the inequalities, we have max ( x − 1 , − x + 1 ) < n < 3 x + 5 . We are also given that x − 1 < n < 3 x + 5 . Since x represents a length, x must be greater than 0. Also, 2 x + 2 , x + 3 and n must be positive. From the triangle inequality, we have n"> 2 x + 2 + x + 3 > n , x+3"> 2 x + 2 + n > x + 3 and 2x+2"> x + 3 + n > 2 x + 2 . This simplifies to n"> 3 x + 5 > n , -x+1"> n > − x + 1 and x-1"> n > x − 1 . Combining these with x − 1 < n < 3 x + 5 , we need to find the intersection of the intervals. Also, we need to consider that x must be large enough for the sides to be positive. Since 0"> x + 3 > 0 and 0"> 2 x + 2 > 0 , we have -3"> x > − 3 and -1"> x > − 1 . Thus -1"> x > − 1 . Also, we need 0"> n > 0 .

Finding the Intersection of the Intervals The possible values of n are given by the intersection of the intervals ( max ( x − 1 , 1 − x ) , 3 x + 5 ) and ( x − 1 , 3 x + 5 ) . Since -1"> x > − 1 , we have -2"> x − 1 > − 2 and 2"> 3 x + 5 > 2 . We also have 1 − x < 2 for -1"> x > − 1 . Thus, max ( x − 1 , 1 − x ) < n < 3 x + 5 and x − 1 < n < 3 x + 5 . Thus, max ( x − 1 , 1 − x ) = ∣ x − 1∣ . If 1"> x > 1 , then x − 1 < n < 3 x + 5 . If − 1 < x < 1 , then 1 − x < n < 3 x + 5 . Thus, ∣ x − 1∣ < n < 3 x + 5 .

Final Answer Therefore, the possible values of n are given by ∣ x − 1∣ < n < 3 x + 5 .


Examples
In construction, when building a triangular structure, the length of one side must be within a certain range, given the lengths of the other two sides. This ensures the structural integrity of the triangle. For example, if you have two sides of lengths 2 x + 2 and x + 3 , the third side n must satisfy the condition ∣ x − 1∣ < n < 3 x + 5 to form a valid triangle.

Answered by GinnyAnswer | 2025-07-03

To determine the possible values of n in the triangle with sides 2 x + 2 , x + 3 , and n , we apply the triangle inequality theorem. The resulting expression for n is x − 1 < n < 3 x + 5 . This provides the range of lengths n can take.
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Answered by Anonymous | 2025-07-04