{x+1}+2x=4[x+1]-6

{}=fractional part..[]=integral part...pls solvve..rates assured.

see, x=[x] +{x}.

now, use this:

(x+1) -[x+1] + 2x = 4[x+1] - 6

3x +7=5[x+1].

3x +7=5[x] + 5.

3x +2 = 5[x].

Now, draw the graphs of y=3x+2 and y=5[x].

the intersection points will give you the desired solutions. You can try hit n trial also.

like 8/3 and 1 are two solutions. I haven't checked for more.

but for clarity try to draw graphs.

If still any doubt exists, do tell us.

 In many of these problems, a useful strategy is try to obtain an equation like {x} = "something" and exploit the fact that 0<={x}<1.

Here LHS =  {x} + 2x = {x} + 2([x]+{x}] = 2[x] +3{x}

So the equation can be written as 3 {x} = 2[x] - 2. or 

So {x} is a rational number of the form n/3 where n is a non-negative even integer. So n = 0 or 2.

n = 0 gives x =1.

n = 2 gives 8/3

as we know...{x}=x-[x]

(x+1)-[x+1]+2x=4[x+1]-6

3x+1=5[x+1]-6

3x+1=5([x]+1)-6

3([x]+{x})=5[x]-2

3{x}=2[x]-2................(1)

as we know..0<={x}<1

0<=3{x}<3........................(2)

on substitutin...(1)

we get..

1<=[x]<5/2...so we conclude that...[x]=1 or 2.

hence from (1)..3{x}=2*1-2=0

or {x}=0 or 2/3.

so

x={x}+[x]...1+0 or 2+2/3  i.e 

1,8/3 are solutions.....

nudge me if i m wrong....