(I assume that the square and curly brackets represent floor and fractional part function respectively)

I am not the right person to help out in such problems but the following is worth a try.
Add and subtract {2x} in the LHS. The equation now becomes 2x-4{2x}=1 or 2x=1+4{2x}. You can now plot a graph to find the number of solutions.

## Comments

Sort by:

TopNewestIf \(\lfloor 2x \rfloor - 3\{2x\} = 1\) then \(3\{2x\} = \lfloor 2x\rfloor - 1\) is an integer, so \(\{2x\} = 0,1/3,2/3\).

If \(\{2x\} = 0\) then \(\lfloor 2x\rfloor - 1 = 0\), so \(\lfloor 2x\rfloor = 1\), and hence \(2x=1\), so \(x=1/2\).

If \(\{2x\} = 1/3\) then \(\lfloor 2x\rfloor - 1 = 1\), so \(\lfloor 2x \rfloor = 2\), and hence \(2x=7/3\), so \(x=7/6\).

If \(\{2x\} = 2/3\) then \(\lfloor 2x\rfloor - 1 = 2\), so \(\lfloor 2x \rfloor = 3\), and hence \(2x=11/3\), so \(x=11/6\).

Log in to reply

Thanks

Log in to reply

(I assume that the square and curly brackets represent floor and fractional part function respectively)

I am not the right person to help out in such problems but the following is worth a try. Add and subtract {2x} in the LHS. The equation now becomes 2x-4{2x}=1 or 2x=1+4{2x}. You can now plot a graph to find the number of solutions.

Log in to reply

Thanks.

Log in to reply