Jej pls help i need my brother help

Answer:
[tex]\mathsf {\frac{17}{10}}[/tex]
Step-by-step explanation:
[tex]\mathsf {2\frac{2}{3} + \frac{14}{5} + ? = \frac{43}{6} }[/tex]
[tex]\mathsf {\frac{8}{3} + \frac{14}{5} + ? = \frac{43}{6} }[/tex]
Take the LCM (common denominator) to be 30.
[tex]\mathsf {\frac{80}{30} + \frac{84}{30} + ? = \frac{215}{30} }[/tex]
Bring the known values to the RHS.
[tex]\mathsf {? = \frac{215}{30} - \frac{164}{30} }[/tex]
[tex]\mathsf {? = \frac{51}{30}}[/tex]
[tex]\implies \mathsf {\frac{17}{10}}[/tex]
Let that be x
Now
[tex]\\ \rm\rightarrowtail \dfrac{8}{3}+\dfrac{14}{5}+x=\dfrac{43}{6}[/tex]
[tex]\\ \rm\rightarrowtail \dfrac{40+42}{15}+x=\dfrac{43}{6}[/tex]
[tex]\\ \rm\rightarrowtail \dfrac{82}{15}+x=\dfrac{43}{6}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{43}{6}-\dfrac{82}{15}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{215-164}{30}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{51}{30}[/tex]
[tex]\\ \rm\rightarrowtail x=\dfrac{17}{10}[/tex]
[tex]\\ \rm\rightarrowtail x=1\dfrac{7}{10}[/tex]