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Question Number 92778 by unknown last updated on 09/May/20

Commented by prakash jain last updated on 09/May/20

$$\lfloor 3x\rfloor -2-(\lfloor 2x\rfloor -1)=2x-6$$$$\lfloor 3x\rfloor -\lfloor 2x\rfloor =2x-5$$$$\mathrm{since}\mathrm{LHS}\mathrm{is}\mathrm{an}\mathrm{integer}\mathrm{RHS}$$$$\mathrm{must}\mathrm{be}\mathrm{integer}$$$$\Rightarrow x\mathrm{must}\mathrm{be}\mathrm{of}\mathrm{form}n\mathrm{or}\frac{n}{2}$$$$x\mathrm{is}\mathrm{of}\mathrm{form}n$$$$3n-2n=2n+7$$$$n=5$$$$x\mathrm{is}\mathrm{of}\mathrm{form}\frac{n}{2}$$$$\lfloor \frac{3n}{2}\rfloor -\lfloor 2\frac{n}{2}\rfloor =\frac{2n}{2}-5$$$$\lfloor \frac{3n}{2}\rfloor -n=n-5$$$$\lfloor \frac{3n}{2}\rfloor =2n-5\left(A\right)$$$$\mathrm{Now}n\mathrm{must}\mathrm{be}\mathrm{of}\mathrm{form}2k+1$$$$\mathrm{other}x\mathrm{will}\mathrm{be}\mathrm{of}\mathrm{form}n\mathrm{for}\mathrm{which}$$$$\mathrm{we}\mathrm{have}\mathrm{already}\mathrm{solved}.$$$$n=2k+1$$$$\lfloor \frac{3(2k+1)}{2}\rfloor =2(2k+1)-5$$$$\lfloor 3k+1+\frac{1}{2}\rfloor =2(2k+1)-5$$$$3k+1=4k-3$$$$\Rightarrow k=4\mathrm{or}n=9\mathrm{or}x=\frac{9}{2}$$$$\mathrm{two}\mathrm{solution}$$$$x=5\mathrm{or}x=\frac{9}{2}$$

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