Question Number 87093 by M±th+et£s last updated on 02/Apr/20
$$\lfloor\frac{{x}−\mathrm{1}}{\mathrm{4}}\rfloor+\lfloor\frac{{x}−\mathrm{2}}{\mathrm{3}}\rfloor=\lfloor\frac{{x}−\mathrm{3}}{\mathrm{2}}\rfloor \\ $$
Commented by M±th+et£s last updated on 02/Apr/20
$${slove}\:{the}\:{equation} \\ $$
Commented by MJS last updated on 02/Apr/20
$$\mathrm{similar}\:\mathrm{to}\:\mathrm{qu}.\:\mathrm{86009} \\ $$
Commented by M±th+et£s last updated on 02/Apr/20
$${yes}\:{sir}\:{thank}\:{you} \\ $$
Commented by MJS last updated on 02/Apr/20
$$\mathrm{I}\:\mathrm{get}\:\mathrm{for}\:{x}\in\mathbb{Z} \\ $$$${x}\in\left\{−\mathrm{10},\:−\mathrm{7},\:−\mathrm{6},\:−\mathrm{4},\:−\mathrm{3},\:−\mathrm{2},\:−\mathrm{1},\:\mathrm{0},\:\mathrm{1},\:\mathrm{3},\:\mathrm{4},\:\mathrm{7}\right\} \\ $$$$\mathrm{for}\:{x}\in\mathbb{R}\:\mathrm{these}\:\mathrm{intervals}: \\ $$$$\left[−\mathrm{10},\:−\mathrm{9}\left[\right.\right. \\ $$$$\left[−\mathrm{7},\:−\mathrm{5}\left[\right.\right. \\ $$$$\left[−\mathrm{4},\:\mathrm{2}\left[\right.\right. \\ $$$$\left[\mathrm{3},\:\mathrm{4}\left[\right.\right. \\ $$$$\left[\mathrm{7},\:\mathrm{8}\left[\right.\right. \\ $$
Commented by M±th+et£s last updated on 02/Apr/20
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by M±th+et£s last updated on 02/Apr/20
$$\:{can}\:{you}\:{show}\:{your}\:{work}\:{sir} \\ $$