Respuesta :
since we know R is the midpoint, then QR = RS = 5x - 7.
[tex]\bf \underset{\leftarrow \qquad \textit{\large 96}\qquad \to }{\boxed{Q}\stackrel{5x-7}{\rule[0.35em]{10em}{0.25pt}} R\stackrel{\underline{5x-7}}{\rule[0.35em]{10em}{0.25pt}\boxed{S}}} \\\\\\ (5x-7)+(5x-7)=96\implies 10x-14=96\implies 10x=110 \\\\\\ x=\cfrac{110}{10}\implies x=11[/tex]
Answer: The required value of x is 11.
Step-by-step explanation: Given that R is the midpoint of the line segment QS, where
QR=5x-7 and QS=96.
We are to find the value of x.
Since the point R is the midpoint of the segment QS, so we must have
[tex]QR=SR=\dfrac{1}{2}QR\\\\\\\Rightarrow QR=\dfrac{1}{2}QS\\\\\\\Rightarrow 5x-7==\dfrac{1}{2}\times96\\\\\Rightarrow 5x-7=48\\\\\Rightarrow 5x=48+7\\\\\Rightarrow 5x=55\\\\\Righatrrow x=\dfrac{55}{5}\\\\\Rightarrow x=11.[/tex]
Thus, the required value of x is 11.
