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3x-5-x-9-




Question Number 139076 by bramlexs22 last updated on 22/Apr/21
⌈ 3x + 5⌊ x ⌋ ⌉ = 9
$$\lceil\:\mathrm{3x}\:+\:\mathrm{5}\lfloor\:\mathrm{x}\:\rfloor\:\rceil\:=\:\mathrm{9} \\ $$
Answered by mathmax by abdo last updated on 22/Apr/21
⇒[3x]+5[x]=9   let [x]=n ⇒n≤x<n+1 ⇒3n≤3x<3n+3  if 3n≤3x<3n+1 ⇒[3x]=3n  and e⇒3n+5n=9 ⇒8n=9  impossible  if 3n+1≤3x<3n+2 ⇒[3x]=3n+1  and e⇒3n+1+5n=9 ⇒8n=8  ⇒n=1 ⇒4≤3x<5 ⇒(4/3)≤x<(5/3)  if  3n+2≤3x<3n+3 ⇒[3x]=3n+2 and e⇒3n+2+5n=9 ⇒  8n=7  impossible so the set of solution is S=[(4/3),(5/3)[
$$\Rightarrow\left[\mathrm{3x}\right]+\mathrm{5}\left[\mathrm{x}\right]=\mathrm{9}\:\:\:\mathrm{let}\:\left[\mathrm{x}\right]=\mathrm{n}\:\Rightarrow\mathrm{n}\leqslant\mathrm{x}<\mathrm{n}+\mathrm{1}\:\Rightarrow\mathrm{3n}\leqslant\mathrm{3x}<\mathrm{3n}+\mathrm{3} \\ $$$$\mathrm{if}\:\mathrm{3n}\leqslant\mathrm{3x}<\mathrm{3n}+\mathrm{1}\:\Rightarrow\left[\mathrm{3x}\right]=\mathrm{3n}\:\:\mathrm{and}\:\mathrm{e}\Rightarrow\mathrm{3n}+\mathrm{5n}=\mathrm{9}\:\Rightarrow\mathrm{8n}=\mathrm{9}\:\:\mathrm{impossible} \\ $$$$\mathrm{if}\:\mathrm{3n}+\mathrm{1}\leqslant\mathrm{3x}<\mathrm{3n}+\mathrm{2}\:\Rightarrow\left[\mathrm{3x}\right]=\mathrm{3n}+\mathrm{1}\:\:\mathrm{and}\:\mathrm{e}\Rightarrow\mathrm{3n}+\mathrm{1}+\mathrm{5n}=\mathrm{9}\:\Rightarrow\mathrm{8n}=\mathrm{8} \\ $$$$\Rightarrow\mathrm{n}=\mathrm{1}\:\Rightarrow\mathrm{4}\leqslant\mathrm{3x}<\mathrm{5}\:\Rightarrow\frac{\mathrm{4}}{\mathrm{3}}\leqslant\mathrm{x}<\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{if}\:\:\mathrm{3n}+\mathrm{2}\leqslant\mathrm{3x}<\mathrm{3n}+\mathrm{3}\:\Rightarrow\left[\mathrm{3x}\right]=\mathrm{3n}+\mathrm{2}\:\mathrm{and}\:\mathrm{e}\Rightarrow\mathrm{3n}+\mathrm{2}+\mathrm{5n}=\mathrm{9}\:\Rightarrow \\ $$$$\mathrm{8n}=\mathrm{7}\:\:\mathrm{impossible}\:\mathrm{so}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{S}=\left[\frac{\mathrm{4}}{\mathrm{3}},\frac{\mathrm{5}}{\mathrm{3}}\left[\right.\right. \\ $$
Answered by mr W last updated on 22/Apr/21
say x=n+f with 0≤f<1  9=⌈3x⌉+5⌊x⌋=⌈3x⌉+5n≥8n ⇒n≤(9/8)  9=⌈3x⌉+5⌊x⌋=⌈3x⌉+5n≤8n+3 ⇒n≥(3/4)  ⇒n=1  ⌈3x⌉=4  3<3x≤4  ⇒1<x≤(4/3)
$${say}\:{x}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\mathrm{9}=\lceil\mathrm{3}{x}\rceil+\mathrm{5}\lfloor{x}\rfloor=\lceil\mathrm{3}{x}\rceil+\mathrm{5}{n}\geqslant\mathrm{8}{n}\:\Rightarrow{n}\leqslant\frac{\mathrm{9}}{\mathrm{8}} \\ $$$$\mathrm{9}=\lceil\mathrm{3}{x}\rceil+\mathrm{5}\lfloor{x}\rfloor=\lceil\mathrm{3}{x}\rceil+\mathrm{5}{n}\leqslant\mathrm{8}{n}+\mathrm{3}\:\Rightarrow{n}\geqslant\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{n}=\mathrm{1} \\ $$$$\lceil\mathrm{3}{x}\rceil=\mathrm{4} \\ $$$$\mathrm{3}<\mathrm{3}{x}\leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{1}<{x}\leqslant\frac{\mathrm{4}}{\mathrm{3}} \\ $$

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