What is the concept for solving equations with integer parts?

[mtayab1994]

Homework Statement

2*[x]-3*[3x]+7=0

Homework Equations

solve for x when [x] is the integer part of the number.

The Attempt at a Solution

To solve it i removed the brackets and got x=1 and then i found the integer part of 1 is 1.

Is it correct? If not is there a concept to solving it?

 

[phyzguy]

What if x = 1.07? Will this be a solution to the equation?

 

[mtayab1994]

what do u mean by what if x=1.07

The integer part of 1.07 is 1 right? so that can be a solution as well and i also thought about have it as x= [1,2[

 

[HallsofIvy]

Don't forget that $[3x]\ne 3[x]$. For example, if 1.5, then [x]= 1 but 3x= 4.5 so [3x]= 4, not 3. With x= 1.5, 2[x]- 3[3x]+ 7= 2(1)- 3(4)+ 7= 2- 12+ 7= -3, not 0.

 

[mtayab1994]

yea thanks for the tip but I'm getting no where with this one at all.

 

[phyzguy]

What are the largest and smallest values of x that will be solutions to the equation? The solution interval is not [1,2], like you said earlier, since, as HallsofIvy showed, 1.5 is not a solution.

 

[Dickfore]

You need to use general intervals. Remember that:
$$[3 x] = N \in \mathbb{Z}, \; N \le 3 x < N + 1 \Leftrightarrow \frac{N}{3} \le x < \frac{N + 1}{3}$$

So, consider the following 3 families of intervals:
First family:

$$N = 3 k, k \le x < k + \frac{1}{3}, \; \mathbb{Z}$$
Then you have:
$$[x] = k, [3 x] = N = 3 k$$
and your equation reads:
$$2 k - 3 (3 k) + 7 = 0$$

Second family:
$$N = 3 k + 1, k + \frac{1}{3} \le x < k + \frac{2}{3}$$
Then, you have:
$$[x] = k, [3 x] = N = 3 k + 1$$
and your equation reads:
$$2 k - 3 ( 3 k + 1) + 7 = 0$$

Third family:
$$N = 3 k + 2, k + \frac{2}{3} \le x < k + 1$$
Then you have:
$$[x] = k, [3 x] = N = 3 k + 2$$
and your equation reads:
$$2 k - 3 (3 k + 2) + 7 = 0$$

Thus, you get three cases that exhaust the whole set of real numbers and for each case you get an equation. The solution to those equations has to be an integer!

 

[phyzguy]

Isn't it true that only the first family has a solution?

 

[Dickfore]

It seems that way, but I wanted the op to check that themselves.