Question

In segment addition how do I find x when DE = x - 2, EF = x + 5, and DF = 9

Answer 1

$\bf D\underset{\leftarrow DF=9\to \qquad }{\stackrel{x-2}{\rule[0.35em]{10em}{0.25pt}}E\stackrel{x+5}{\rule[0.35em]{15em}{0.25pt}}}F \\\\\\ DF=DE+EF\implies \stackrel{DF}{9}=\stackrel{DE}{(x-2)}+\stackrel{EF}{(x+5)}\implies 9=2x+3 \\\\\\ 6=2x\implies \cfrac{6}{2}=x\implies 3=x$