Question

use the rectangle diagram.
a. write and solve an inequality to find the values of ( x ) for which the perimeter of the rectangle is less than 120.
b. based on your answer to part a, are there any values that can be eliminated from the solution set? explain.

a. write an inequality. choose the correct answer
( \bigcirc ) a. ( x + x + 4 lt 120 )
( \bigcirc ) b. ( 2x + 2(x + 4) gt 120 )
( \bigcirc ) c. ( 2x + 2(x + 4) lt 120 )
( \bigcirc ) d. ( x + x + 4 gt 120 )

Explanation:

Step1: Calculate the perimeter formula

The perimeter \(P\) of a rectangle is \(P = 2\times(\text{length}+\text{width})\). Here, length \(=x + 4\) and width \(=x\). So \(P=2x+2(x + 4)\).

Step2: Set up the inequality

Since the perimeter is less than 120, the inequality is \(2x+2(x + 4)<120\).

Step3: Solve the inequality

Expand: \(2x+2x+8<120\).
Combine like - terms: \(4x+8<120\).
Subtract 8 from both sides: \(4x<120 - 8\), so \(4x<112\).
Divide both sides by 4: \(x<28\).

Answer:

a. The correct inequality is C. \(2x + 2(x + 4)<120\), and the solution of the inequality \(2x+2(x + 4)<120\) is \(x<28\).
b. Yes. Since \(x\) represents a dimension (width) of a rectangle, \(x>0\). So values of \(x\) that are non - positive (\(x\leq0\)) can be eliminated from the solution set. Because a side length of a rectangle cannot be negative or zero.