Question Number 95309 by i jagooll last updated on 24/May/20
![if y = [ 2x+5 ] = 3[x−4] then [ 3x+y ] = ?](Q95309.png)
$$\mathrm{if}\:\mathrm{y}\:=\:\left[\:\mathrm{2x}+\mathrm{5}\:\right]\:=\:\mathrm{3}\left[\mathrm{x}−\mathrm{4}\right]\: \\ $$$$\mathrm{then}\:\left[\:\mathrm{3x}+\mathrm{y}\:\right]\:=\:?\: \\ $$
Answered by mr W last updated on 24/May/20
![let u=x−4=n+f with 0≤f<1 2x+5=2(4+n+f)+5=2(4+n)+5+2f case 0≤f<(1/2): y=[2x+5]=2(4+n)+5+[2f]=13+2n=3n ⇒n=13 ⇒x=4+n+f=17+f i.e. 17≤x<17.5 [3x+y]=51+[3f]+39=90+[3f[=90 or 91 case (1/2)≤f<1: y=[2x+5]=2(4+n)+5+[2f]=14+2n=3n ⇒n=14 ⇒x=4+n+f=18+f i.e. 18.5≤x<19 [3x+y]=54+[3f]+42=96+[3f[=97 or 98 ⇒[3x+y]=90, 91, 97, 98.](Q95318.png)
$${let}\:{u}={x}−\mathrm{4}={n}+{f}\:{with}\:\mathrm{0}\leqslant{f}<\mathrm{1} \\ $$$$\mathrm{2}{x}+\mathrm{5}=\mathrm{2}\left(\mathrm{4}+{n}+{f}\right)+\mathrm{5}=\mathrm{2}\left(\mathrm{4}+{n}\right)+\mathrm{5}+\mathrm{2}{f} \\ $$$${case}\:\mathrm{0}\leqslant{f}<\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$${y}=\left[\mathrm{2}{x}+\mathrm{5}\right]=\mathrm{2}\left(\mathrm{4}+{n}\right)+\mathrm{5}+\left[\mathrm{2}{f}\right]=\mathrm{13}+\mathrm{2}{n}=\mathrm{3}{n} \\ $$$$\Rightarrow{n}=\mathrm{13} \\ $$$$\Rightarrow{x}=\mathrm{4}+{n}+{f}=\mathrm{17}+{f} \\ $$$${i}.{e}.\:\mathrm{17}\leqslant{x}<\mathrm{17}.\mathrm{5} \\ $$$$\left[\mathrm{3}{x}+{y}\right]=\mathrm{51}+\left[\mathrm{3}{f}\right]+\mathrm{39}=\mathrm{90}+\left[\mathrm{3}{f}\left[=\mathrm{90}\:{or}\:\mathrm{91}\right.\right. \\ $$$$ \\ $$$${case}\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{f}<\mathrm{1}: \\ $$$${y}=\left[\mathrm{2}{x}+\mathrm{5}\right]=\mathrm{2}\left(\mathrm{4}+{n}\right)+\mathrm{5}+\left[\mathrm{2}{f}\right]=\mathrm{14}+\mathrm{2}{n}=\mathrm{3}{n} \\ $$$$\Rightarrow{n}=\mathrm{14} \\ $$$$\Rightarrow{x}=\mathrm{4}+{n}+{f}=\mathrm{18}+{f} \\ $$$${i}.{e}.\:\mathrm{18}.\mathrm{5}\leqslant{x}<\mathrm{19} \\ $$$$\left[\mathrm{3}{x}+{y}\right]=\mathrm{54}+\left[\mathrm{3}{f}\right]+\mathrm{42}=\mathrm{96}+\left[\mathrm{3}{f}\left[=\mathrm{97}\:{or}\:\mathrm{98}\right.\right. \\ $$$$ \\ $$$$\Rightarrow\left[\mathrm{3}{x}+{y}\right]=\mathrm{90},\:\mathrm{91},\:\mathrm{97},\:\mathrm{98}. \\ $$