Let `{x}` and `[x]` denotes the fraction fractional and integral part of a real number `(x)`, respectively. Solve `|2x-1+=3[x]+2[x]`.

Explanation

Case I `2x-1ge0` or `xge1/2`
Then given equation convert to
`2x-1=3[x]+2{x}`………
`:'x=[x]+{x}` …..ii
From Eq. I and ii we get
`2([x]+{x})-1=3[x]+2{x}`
`:.[x]=-1`
`:.-1lexlt0`
No solution `[:'x ge1/2]`
Case II `2x-1lt0` or `x lt1/2`
Then given equation reduces to
`1-2x=3[x]+2{x}`.....iii
`:'[x=[x]+{x}`.....iv
From Eqs (iii) and (iv) we get
`1-2([x]+{x})=3[x]+2{x}`
`implies1-5[x]=4{x}`
`:.{x}=(1=5[x])/4`.......v
Now `0le{x}lt1`
`implies0le(1-5[x])/4lt1`
`implies0le1-5[x]gt-4` ltbRgt `implies1ge5[x]ge-3` or `-3/5lt[x]le1/5`
`:.[x]=0`
From Eq. (v) `{x}=1/4`
`:.x=0+1/4=1/4`