Solve the following equations.
\lfloor x-\dfrac{2}{3} \rfloor=4
\lfloor E\left(x\right) \rfloor=a if and only if a\leqslant E\left(x\right)\lt a+1.
Therefore:
\lfloor x-\dfrac{2}{3} \rfloor=4
If and only if:
4\leqslant x-\dfrac{2}{3}\lt5
Adding \dfrac{2}{3} to all sides gives:
4+\dfrac{2}{3} \leqslant x\lt5+\dfrac{2}{3}
\dfrac{14}{3}\leqslant x\lt\dfrac{17}{3}
x is a solution if and only if \dfrac{14}{3}\leqslant x\lt\dfrac{17}{3}.
\lfloor 2x+\dfrac{5}{6} \rfloor=3
\lfloor E\left(x\right) \rfloor=a if and only if a\leqslant E\left(x\right)\lt a+1.
Therefore:
\lfloor 2x+\dfrac{5}{6} \rfloor=3
If and only if:
3\leqslant 2x+\dfrac{5}{6}\lt4
Subtracting \dfrac{5}{6} from all sides gives:
3-\dfrac{5}{6} \leqslant 2x\lt4-\dfrac{5}{6}
\dfrac{13}{6}\leqslant 2x\lt\dfrac{19}{6}
Dividing all sides by 2 gives:
\dfrac{13}{12}\leqslant x\lt\dfrac{19}{12}
x is a solution if and only if \dfrac{13}{12}\leqslant x\lt\dfrac{19}{12}.
\lfloor 4x-\dfrac{1}{2} \rfloor=7
\lfloor E\left(x\right) \rfloor=a if and only if a\leqslant E\left(x\right)\lt a+1.
Therefore:
\lfloor 4x-\dfrac{1}{2} \rfloor=7
If and only if:
7\leqslant 4x-\dfrac{1}{2}\lt8
Adding \dfrac{1}{2} to all sides gives:
7+\dfrac{1}{2} \leqslant 4x\lt8+\dfrac{1}{2}
\dfrac{15}{2}\leqslant 4x\lt\dfrac{17}{2}
Dividing all sides by 4 gives:
\dfrac{15}{8}\leqslant x\lt\dfrac{17}{8}
x is a solution if and only if \dfrac{15}{8}\leqslant x\lt\dfrac{17}{8}.
\lfloor 3x-5 \rfloor=1
\lfloor E\left(x\right) \rfloor=a if and only if a\leqslant E\left(x\right)\lt a+1.
Therefore:
\lfloor 3x-5 \rfloor=1
If and only if:
1\leqslant 3x-5\lt2
Adding 5 to all sides gives:
6 \leqslant 3x\lt7
Dividing all sides by 3 gives:
2\leqslant x\lt\dfrac{7}{3}
x is a solution if and only if 2\leqslant x\lt\dfrac{7}{3}.
\lfloor \dfrac{1}{3}x+\dfrac{5}{6} \rfloor=2
\lfloor E\left(x\right) \rfloor=a if and only if a\leqslant E\left(x\right)\lt a+1.
Therefore:
\lfloor \dfrac{1}{3}x+\dfrac{5}{6} \rfloor=2
If and only if:
2\leqslant \dfrac{1}{3}x+\dfrac{5}{6}\lt3
Subtracting \dfrac{5}{6} from all sides gives:
2-\dfrac{5}{6} \leqslant \dfrac{1}{3}x\lt3-\dfrac{5}{6}
\dfrac{7}{6} \leqslant \dfrac{1}{3}x\lt\dfrac{13}{6}
Multiplying all sides by 3 gives:
\dfrac{7}{2}\leqslant x\lt\dfrac{13}{2}
x is a solution if and only if \dfrac{7}{2}\leqslant x\lt\dfrac{13}{2}.
\lfloor \dfrac{2}{5}x-\dfrac{7}{6} \rfloor=5
\lfloor E\left(x\right) \rfloor=a if and only if a\leqslant E\left(x\right)\lt a+1.
Therefore:
\lfloor \dfrac{2}{5}x-\dfrac{7}{6} \rfloor=5
If and only if:
5\leqslant \dfrac{2}{5}x-\dfrac{7}{6}\lt6
Adding \dfrac{7}{6} to all sides gives:
5+\dfrac{7}{6} \leqslant \dfrac{2}{5}x\lt6+\dfrac{7}{6}
\dfrac{37}{6} \leqslant \dfrac{2}{5}x\lt\dfrac{43}{6}
Multiplying all sides by \dfrac{5}{2} gives:
\dfrac{185}{12}\leqslant x\lt\dfrac{215}{12}
x is a solution if and only if \dfrac{185}{12}\leqslant x\lt\dfrac{215}{12}.
\lfloor 3x+\dfrac{1}{7} \rfloor=0
\lfloor E\left(x\right) \rfloor=a if and only if a\leqslant E\left(x\right)\lt a+1.
Therefore:
\lfloor 3x+\dfrac{1}{7} \rfloor=0
If and only if:
0\leqslant 3x+\dfrac{1}{7}\lt1
Subtracting \dfrac{1}{7} from all sides gives:
0-\dfrac{1}{7} \leqslant 3x\lt1-\dfrac{1}{7}
-\dfrac{1}{7} \leqslant 3x\lt\dfrac{6}{7}
Dividing all sides by 3 gives:
-\dfrac{1}{21}\leqslant x\lt\dfrac{2}{7}
x is a solution if and only if -\dfrac{1}{21}\leqslant x\lt\dfrac{2}{7}.