Question

geometry
part 2: segment addition (honors)
solve for x.

1. u ---3--- t ---x-1--- s ---2--- r

\t\t\t\t\t\t\t 2x - 7

Explanation:

Step1: Apply Segment Addition Postulate

The total length from \( U \) to the end (let's say \( V \), though labeled as the arrow) should be the sum of \( UT \), \( TS \), and \( SR \). Wait, actually, looking at the diagram: \( UT = 3 \), \( TS = x - 1 \), \( SR = 2 \), and the total from \( U \) to the arrow (let's say \( UR \) but the lower segment is \( 2x - 7 \) from \( U \) to the arrow. Wait, maybe the lower segment is \( U \) to the arrow, and the upper segments are \( U \) to \( T \) (3), \( T \) to \( S \) (\( x - 1 \)), \( S \) to \( R \) (2). But the lower segment is \( U \) to the arrow, which should equal \( UT + TS + SR \)? Wait, no, maybe the arrow is the same as \( R \)? Wait, the diagram: \( U \)---(3)---\( T \)---(\( x - 1 \))---\( S \)---(2)---\( R \), and below, from \( U \) to the arrow (which is at the end of \( R \)?) is \( 2x - 7 \). So by Segment Addition Postulate, \( UT + TS + SR = \) the length from \( U \) to \( R \) (the arrow). So \( 3 + (x - 1) + 2 = 2x - 7 \).

Simplify left side: \( 3 + x - 1 + 2 = x + 4 \). So equation: \( x + 4 = 2x - 7 \).

Step2: Solve for \( x \)

Subtract \( x \) from both sides: \( 4 = x - 7 \).

Add 7 to both sides: \( x = 11 \).

Answer:

\( x = 11 \)