We will solve this problem first, by solving the inequality in the left hand side and next the inequality on the right hand side.
In the left hand side, we have
[tex]-\frac{2}{5}\le\frac{x+4}{3}[/tex]
If we move 3 to the left hand side, we obtain
[tex]-\frac{2}{5}\cdot3\le x+4[/tex]
which is equal to
[tex]-\frac{6}{5}\le x+4[/tex]
Now, if we move 4 to the left hand side as -4, we have
[tex]\begin{gathered} -\frac{6}{5}-4\le x \\ -\frac{6}{5}-\frac{20}{5}\le x \\ \frac{-6-20}{5}\le x \\ -\frac{26}{5}\le x \end{gathered}[/tex]
Now, in the right hand side, we have
[tex]\frac{x+4}{3}and if we move 3 to the right hand side, we obtain[tex]x+4<3(x+5)[/tex]
we must note that, since 3 is positive, it doesnt flipt the inequality sign. Then, we obtain
[tex]x+4<3x+15[/tex]
Now, if we move x to the right hand side we have
[tex]\begin{gathered} 4<3x-x+15 \\ 4<2x+15 \end{gathered}[/tex]
and finally, we have
[tex]\begin{gathered} 4-15<2x \\ -11<2x \\ \frac{-11}{2}In summary, we have the following conditions:[tex]-\frac{26}{5}\le x[/tex]
and
[tex]\frac{-11}{2}
and we must choose one of them. We can see that
[tex]\begin{gathered} -\frac{11}{2}<-\frac{26}{5} \\ \text{because} \\ -5.5<-5.2 \end{gathered}[/tex]
Therefore, the answer which fulfil both conditions is
[tex]-\frac{26}{5}\le x[/tex]
