Solve the following inequalities.
\lfloor x+\dfrac{4}{5} \rfloor \lt \dfrac{11}4
\lfloor E\left(x\right) \rfloor\lt a if and only if E\left(x\right)\lt \lfloor a \rfloor+1.
Therefore:
\lfloor x+\dfrac{4}{5} \rfloor \lt \dfrac{11}4
If and only if:
x+\dfrac{4}{5} \lt 3
Subtracting \dfrac{4}{5} from both sides gives:
x \lt 3-\dfrac{4}{5}
x \lt \dfrac{11}{5}
x is a solution if and only if x \lt \dfrac{11}{5}.
\lfloor 2x+\dfrac{1}{4} \rfloor \lt \dfrac{3}7
\lfloor E\left(x\right) \rfloor\lt a if and only if E\left(x\right)\lt \lfloor a \rfloor+1.
Therefore:
\lfloor 2x+\dfrac{1}{4} \rfloor\lt \dfrac{3}7
If and only if:
2x+\dfrac{1}{4} \lt 1
Subtracting \dfrac{1}{4} from both sides gives:
2x \lt 1-\dfrac{1}{4}
2x \lt \dfrac{3}{4}
Dividing both sides by 2 gives:
x \lt \dfrac{3}{8}
x is a solution if and only if x \lt \dfrac{3}{8}.
\lfloor \dfrac{1}{4}x-\dfrac{2}{3} \rfloor \lt \dfrac{7}3
\lfloor E\left(x\right) \rfloor\lt a if and only if E\left(x\right)\lt \lfloor a \rfloor+1.
Therefore:
\lfloor \dfrac{1}{4}x-\dfrac{2}{3} \rfloor \lt \dfrac{7}3
If and only if:
\dfrac{1}{4}x-\dfrac{2}{3} \lt 3
Adding \dfrac{2}{3} to both sides gives:
\dfrac{1}{4}x \lt 3+\dfrac{2}{3}
\dfrac{1}{4}x \lt \dfrac{11}{3}
Multiplying both sides by 4 gives:
x \lt \dfrac{44}{3}
x is a solution if and only if x \lt \dfrac{44}{3}.
\lfloor \dfrac{3}{4}x-\dfrac{1}{5} \rfloor \lt -\dfrac{13}6
\lfloor E\left(x\right) \rfloor\lt a if and only if E\left(x\right)\lt \lfloor a \rfloor+1.
Therefore:
\lfloor \dfrac{3}{4}x-\dfrac{1}{5} \rfloor \lt -\dfrac{13}6
If and only if:
\dfrac{3}{4}x-\dfrac{1}{5} \lt -2
Adding \dfrac{1}{5} to both sides gives:
\dfrac{3}{4}x \lt -2+\dfrac{1}{5}
\dfrac{3}{4}x \lt -\dfrac{9}{5}
Multiplying both sides by \dfrac{4}{3} :
x \lt -\dfrac{12}{5}
x is a solution if and only if x \lt -\dfrac{12}{5}.
\lfloor x-\dfrac{2}{3} \rfloor \gt \dfrac{5}6
\lfloor E\left(x\right) \rfloor\gt a if and only if E\left(x\right)\geq \lfloor a \rfloor+1.
Therefore:
\lfloor x-\dfrac{2}{3} \rfloor \gt \dfrac{5}6
If and only if:
x-\dfrac{2}{3} \geqslant1
Adding \dfrac{2}{3} to both sides gives:
x \geqslant 1+\dfrac{2}{3}
x \geqslant \dfrac{5}{3}
x is a solution if and only if x \geqslant \dfrac{5}{3}.
\lfloor 3x-\dfrac{1}{5} \rfloor\gt \dfrac{9}2
\lfloor E\left(x\right) \rfloor\gt a if and only if E\left(x\right)\geq \lfloor a \rfloor+1.
Therefore:
\lfloor 3x-\dfrac{1}{5} \rfloor \gt \dfrac{9}2
If and only if:
3x-\dfrac{1}{5} \geqslant5
Adding \dfrac{1}{5} to both sides gives:
3x \geqslant 5+\dfrac{1}{5}
3x \geqslant \dfrac{26}{5}
Dividing both sides by 3 gives:
x \geqslant \dfrac{26}{15}
x is a solution if and only if x \geqslant \dfrac{26}{15}.
\lfloor x+4 \rfloor\gt 2
\lfloor E\left(x\right) \rfloor\gt a if and only if E\left(x\right)\geq \lfloor a \rfloor+1.
Therefore:
\lfloor x+4 \rfloor\gt 2
If and only if:
x+4 \geqslant3
Subtracting 4 from both sides gives:
x\geqslant-1
x is a solution if and only if x\geqslant-1.