Question:
Solve the equation
$$\left\{ \frac {3x + 5}{x + 2} \right\}+ \left\lfloor \frac{3x + 2}{x + 1} \right\rfloor = \frac{25}{9}$$
where $\lfloor a \rfloor$, $\{a\}$
represents the greatest integer $≤a$, respectively fractional part of $a$.
My idea:
Ok, so first I wrote $\frac{3x + 5}{x + 2}= 2+ \frac{x+1}{x+2}$
and $ \frac{3x + 2}{x + 1} = 2+ \frac{x}{x+1}$. From here I got $2$ cases:
- If $x$ is positive: then we got $x<x+1$ and $x+1<x+2$, so:
$$\begin{align}&\left\{\frac{3x + 5}{x + 2} \right\}= \left\{ 2+ \frac{x+1}{x+2} \right\}= \frac{x+1}{x+2}\\&\left\lfloor \frac{3x + 2}{x + 1} \right\rfloor= \left\lfloor 2+ \frac{x}{x+1} \right\rfloor =2\end{align}$$
So we get that: $$\frac{x+1}{x+2} +2 = \frac{25}{9}\:\Longrightarrow\:x= \frac{5}{2}$$
- If $x$ is negative: similary we get $$\left\{ \frac{3x + 5}{x + 2} \right\}= \left\{ 3 – \frac{1}{x+2} \right\}= \frac{x+1}{x+2}$$
We know that $x$ is negative which makes $ 2>\frac{x}{x+1}>1$ and:
$$\left\lfloor \frac{3x + 2}{x + 1} \right\rfloor= \left\lfloor 2+ \frac{x}{x+1} \right\rfloor =3$$
Thus:
$$3+ \frac{x+1}{x+2}= \frac{25}{9}\:\longrightarrow\:x=- \frac{13}{11}$$
Can you tell me if my idea and reasoning are corect? Im not sure of the case were $x$ is negative. Hope one of you can help me! Thank you!
Best Answer
I think mistake is in your cases eg.if $x=-1.9999$ according to your 2nd case you get fractional part equal to $1000$
Simplifying as you have
$$ \{ \frac {x + 1}{x + 2} \}+ \big \lfloor \frac{x}{x + 1} \big \rfloor = \frac{7}{9} $$
Trivially $\big \lfloor \frac{x}{x + 1} \big \rfloor=0$ since RHS<$1$ and fractional part is always positive
$$ 0\le{x\over x+1}\Rightarrow x\in(-\infty,-1)\cup[0,\infty)\\ \ \\ {x\over x+1}\lt1\Rightarrow 0\lt{1\over x+1}\Rightarrow x\in(-1,\infty) $$
The common region being $x\in[0,\infty)$
Now we have for some $n\in\mathbb{N}$ $$ {x+1\over x+2}=n+{7\over9}\\ \ \\ x={5+18n\over 2-9n} $$
We have from the first case than $x\ge0$ which gives $n\in[-{5\over18},{2\over9})$ the only integer in this range being $n=0$ corresponding to $x={5\over2}$