# Math Help - Solve the equation

1. ## Solve the equation

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

2. Hello, james_bond!

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

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

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

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

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

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

And that's it for now . . .

3. 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 $t-1<[t]\leq t$.
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.

4. ## Remarks on Soroban's post

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

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

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

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

And that's it for now . . .

In addition to the comments by wisterville, I have the followings.

At the right-hand side of [1], we do not know that $\frac{x-1}{3}$ is an integer, what we know is $2x+\frac{x-1}{3}$ is an integer exactly.
Letting $x=\frac{13}{7}$, we indeed see that $\frac{x-1}{3}=\frac{2}{7}\not\in\mathbb{Z}$ but we have $2x+\frac{x-1}{3}=6\in\mathbb{Z}$.

We clearly have $[t]\leq t<[t]+1$ for all $t\in\mathbb{R}$.
Then, using the equation you gave with $t=\frac{4x+1}{5}$, we get
$\frac{7x-1}{3}\leq\frac{4x+1}{5}<\frac{7x-1}{3}+1$.
You can solve these inequalities by seperating into two parts.