Answer:
B.) [tex]x<2\frac{11}{15}[/tex]
Step-by-step explanation:
[tex]3.4>x +2/3\\\rule{150}{0.5}\\\\3.4 \rightarrow 3\frac{2}{5} \rightarrow\frac{17}{5}\\\\\frac{17}{5}>x + \frac{2}{3}\\\\\frac{17}{5} - \frac{2}{3} > x + \frac{2}{3}- \frac{2}{3}\\\rule{150}{0.5}\\\\\boxed{\frac{17}{5} - \frac{2}{3}}\\\\\\LCM(3,5): 15\\\\ \frac{2}{3}\rightarrow \frac{2*5}{3*5}\rightarrow\frac{10}{15}\\\\ \frac{17}{5} \rightarrow\frac{17*3}{5*3}\rightarrow\frac{51}{15}\\\\\frac{51}{15} - \frac{10}{15} = \frac{41}{15}\\\rule{150}{0.5}\\ \frac{41}{15} > x\\\\[/tex]
[tex]x < \frac{41}{15}\leftarrow \text{The sides are switched.}\\\\\text{41 can go into 15 2 times with 11 remaining.}\\\\\frac{41}{15}\rightarrow 2\frac{11}{15}\\\\\text{Therefore...} \\\\\boxed{x < 2\frac{11}{15}}[/tex]
Option B should be the correct answer.
Hope this helps.
