Modgifrac Equation

$|2x - 1| = 3 \lfloor {x} \rfloor + 2 \{ x \}$

Since:

$x = \lfloor {x} \rfloor + \{ x \}$

Therefore:

$|2x - 1| = 3 \lfloor {x} \rfloor + 2 \{ x \} \iff |2x - 1| = 2x + \lfloor x \rfloor$

Now, we have two cases:

1.) $2x -1 ≥ 0 \iff x ≥ 0.5$

$2x - 1 = 2x + \lfloor x \rfloor$

$- 1 = \lfloor x \rfloor$

$-1 ≤ x < 0$

However, since x ≥ 0.5 we don't have any solutions in this case.

2.) $2x -1 < 0 \iff x < 0.5$

$1- 2x = 2x + \lfloor x \rfloor$

$1 = 4x + \lfloor x \rfloor$

If x is negative, then it's floor is also negative, so we would have a negative number on the RHS, which cannot be equal the positive LHS, so no solution here.

If 0 ≤ x < 0.5 , then:

$\lfloor x \rfloor = 0$

Therefore:

1 = 4x

x = 0.25

This means, that we only have one solution (x = 0.25).

$\text {Hence, our answer should be: } \boxed {1}$