Solve the equation

• Sep 12th 2008, 02:06 AM
james_bond
Solve the equation
$\displaystyle 3\cdot\left[\frac {4x+1}5\right]=7x-1$ where $\displaystyle [a]$ is the integer part of $\displaystyle a$.
• Sep 12th 2008, 07:22 AM
Soroban
Hello, james_bond!

I haven't solved it yet, but here's an approach.
. . Maybe it will inspire someone . . .

Quote:

$\displaystyle 3\cdot\left[\frac {4x+1}5\right]=7x-1$ .where $\displaystyle [a]$ is the integer part of $\displaystyle a$

We have: .$\displaystyle \left[\frac{4x+1}{5}\right] \;=\;\frac{7x-1}{3} \quad\Rightarrow\quad \left[\frac{4x+1}{5}\right]\;=\;2x + \frac{x-1}{3}\;\;{\color{blue}[1]}$

Since the right side must be an integer, $\displaystyle x-1 \:=\:3k\:\text{, for some integer }k$
. . That is: .$\displaystyle x \:=\:3k+1$

Substitute into [1]: . $\displaystyle \left[\frac{4(3k+1)+1}{5}\right] \;=\;2(3k+1) + \frac{(3k+1)-1}{3}$

. . which simplifies to: .$\displaystyle \left[\frac{12k+5}{5}\right] \;=\;7k+2 \quad\Rightarrow\quad \left[2k+1+\frac{2k}{5}\right] \;=\;7k+2$

And that's it for now . . .

• Sep 12th 2008, 08:12 AM
wisterville
Hello,

Soroban made a mistake since x is not necessarily an integer. Instead, as he pointed out, it is true that (7x-1)/3 is an integer. Thus, x=(3k+1)/7 for some integer k.

You have an inequality $\displaystyle t-1<[t]\leq t$.
In this case, you get (4x+1)/5 - 1 < (7x-1)/3 <= (4x+1)/5 which means -7/23<x<=8/23.
Now, substituting x=(3k+1)/7, you get k=-1, 0.

Bye.
• Sep 24th 2008, 08:42 AM
bkarpuz
Remarks on Soroban's post
Quote:

Originally Posted by Soroban
Hello, james_bond!

I haven't solved it yet, but here's an approach.
. . Maybe it will inspire someone . . .

We have: .$\displaystyle \left[\frac{4x+1}{5}\right] \;=\;\frac{7x-1}{3} \quad\Rightarrow\quad \left[\frac{4x+1}{5}\right]\;=\;2x + \frac{x-1}{3}\;\;{\color{blue}[1]}$

Since the right side must be an integer, $\displaystyle x-1 \:=\:3k\:\text{, for some integer }k$
. . That is: .$\displaystyle x \:=\:3k+1$

Substitute into [1]: . $\displaystyle \left[\frac{4(3k+1)+1}{5}\right] \;=\;2(3k+1) + \frac{(3k+1)-1}{3}$

. . which simplifies to: .$\displaystyle \left[\frac{12k+5}{5}\right] \;=\;7k+2 \quad\Rightarrow\quad \left[2k+1+\frac{2k}{5}\right] \;=\;7k+2$

And that's it for now . . .

At the right-hand side of [1], we do not know that $\displaystyle \frac{x-1}{3}$ is an integer, what we know is $\displaystyle 2x+\frac{x-1}{3}$ is an integer exactly.
Letting $\displaystyle x=\frac{13}{7}$, we indeed see that $\displaystyle \frac{x-1}{3}=\frac{2}{7}\not\in\mathbb{Z}$ but we have $\displaystyle 2x+\frac{x-1}{3}=6\in\mathbb{Z}$.
We clearly have $\displaystyle [t]\leq t<[t]+1$ for all $\displaystyle t\in\mathbb{R}$.
Then, using the equation you gave with $\displaystyle t=\frac{4x+1}{5}$, we get
$\displaystyle \frac{7x-1}{3}\leq\frac{4x+1}{5}<\frac{7x-1}{3}+1$.