1didnt get a word of it :(
10 Answers
62\\\left[\frac{3x-1}{4} \right]+\left[\frac{3x-1}{4} \right]+\left[\frac{6x-2}{4} \right]=\frac{6x+3}{5} \\\left[\frac{3x-1}{4} \right]+\left[\frac{3x-1}{4} \right]=\left\{\frac{6x-2}{5} \right\}+1
So the fraction part is zero or 1
Now can you finish this off?
62except for {} everything else is an integer.. so for two sides to be integers, {} has to be integer as well.. so 0 or 1
1it got deleted ...
i was saying that i see a 4 that should have been a 5 ....
or i am completely missing the point ....
29Bhaiya..i am having some dbts in this question..
One there shud be + sign as shown..
\left[\frac{3x-1}{4} \right] + {\color{blue} \left[\frac{3x+1}{4} \right]}+ \left[\frac{6x-2}{4} \right]
secondly bhaiya how
\frac{6x+3}{5}- \left[\frac{6x-2}{4} \right] = \left\{\frac{6x-2}{5} \right\} + 1
62oh sorry.. i have gone mad you are right philip..
I hshould have used 2X+1 is a multiple of 5
so x=(5k-1)/2
341Use the inequality y-1 ≤[y]≤y to arrive at \frac{11}{18} \le x \le \frac{41}{18}
Also since 3(2x+1) is to be divisible by 5, we must have x an integer of the form 5k+2.
Hence the only solution is x =2
