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Question Number 139076 by bramlexs22 last updated on 22/Apr/21

$$\lceil 3\mathrm{x}+5\lfloor \mathrm{x}\rfloor \rceil =9$$

Answered by mathmax by abdo last updated on 22/Apr/21

$$\Rightarrow \left[3\mathrm{x}\right]+5\left[\mathrm{x}\right]=9\mathrm{let}\left[\mathrm{x}\right]=\mathrm{n}\Rightarrow \mathrm{n}⩽\mathrm{x}<\mathrm{n}+1\Rightarrow 3\mathrm{n}⩽3\mathrm{x}<3\mathrm{n}+3$$$$\mathrm{if}3\mathrm{n}⩽3\mathrm{x}<3\mathrm{n}+1\Rightarrow \left[3\mathrm{x}\right]=3\mathrm{n}\mathrm{and}\mathrm{e}\Rightarrow 3\mathrm{n}+5\mathrm{n}=9\Rightarrow 8\mathrm{n}=9\mathrm{impossible}$$$$\mathrm{if}3\mathrm{n}+1⩽3\mathrm{x}<3\mathrm{n}+2\Rightarrow \left[3\mathrm{x}\right]=3\mathrm{n}+1\mathrm{and}\mathrm{e}\Rightarrow 3\mathrm{n}+1+5\mathrm{n}=9\Rightarrow 8\mathrm{n}=8$$$$\Rightarrow \mathrm{n}=1\Rightarrow 4⩽3\mathrm{x}<5\Rightarrow \frac{4}{3}⩽\mathrm{x}<\frac{5}{3}$$$$\mathrm{if}3\mathrm{n}+2⩽3\mathrm{x}<3\mathrm{n}+3\Rightarrow \left[3\mathrm{x}\right]=3\mathrm{n}+2\mathrm{and}\mathrm{e}\Rightarrow 3\mathrm{n}+2+5\mathrm{n}=9\Rightarrow$$$$8\mathrm{n}=7\mathrm{impossible}\mathrm{so}\mathrm{the}\mathrm{set}\mathrm{of}\mathrm{solution}\mathrm{is}\mathrm{S}=[\frac{4}{3},\frac{5}{3}[$$

Answered by mr W last updated on 22/Apr/21

$$sayx=n+fwith0⩽f<1$$$$9=\lceil 3x\rceil +5\lfloor x\rfloor =\lceil 3x\rceil +5n⩾8n\Rightarrow n⩽\frac{9}{8}$$$$9=\lceil 3x\rceil +5\lfloor x\rfloor =\lceil 3x\rceil +5n⩽8n+3\Rightarrow n⩾\frac{3}{4}$$$$\Rightarrow n=1$$$$\lceil 3x\rceil =4$$$$3<3x⩽4$$$$\Rightarrow 1<x⩽\frac{4}{3}$$

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