Find the real values of $x$ that satisfy the equation $7[x]+23\{x\}=191$ For any real number $x$, $[x]$ denotes the largest integer less than or equal to $x$ (i.e. floor function) and $ \{x\}=x-[x]$ .Then, the number of real solutions of the equation $$7[x]+23\{x\}=191$$ are.
My Attempt:
I used $ \{x\}=x-[x]$
This gives us,
$$23x-16[x]=191$$
as,$16[x]$ is an integer, $23x$ also must be an integer.
How to proceed next?
 A: Hints:


*

*$23x-16[x]=191$ is a nice idea, but $7x+16\{x\}=191$ may be more useful  

*Using $0 \le \{x\} \lt 1$, can you put an upper bound on $7x$? A lower bound? 

*If you knew $[x]$, could you find $\{x\}$ and so $x$?

*How many possible values of $[x]$ are there? Do they all give a value of $x$ which works?
A: Adding the solution here for future references.
$$7[x]+23\{x\}=191$$
We know that, $0\leq23{\{x\}}\lt23$,
i.e. as long as $$168\lt7[x]\leq191$$
We can always find a $23{x}$ (one and only one) in $[[x],[x]+1)$ such that,
$$7[x]+23\{x\}=191$$
This gives us $[x]=25,26,27$
as $[x]=24$ gives $7[x]=168$, this will just lie on the boundary.
So, 3 solutions.
A: Here my attempt...
$\mathrm{7}\left[{x}\right]+\mathrm{23}\left\{{x}\right\}=\mathrm{191} \\ $
$\left[{x}\right]+\frac{\mathrm{23}\left\{{x}\right\}−\mathrm{2}}{\mathrm{7}}=\mathrm{27} \\ $
$\left\{{x}\right\}\in\left\{\frac{\mathrm{2}}{\mathrm{23}},\frac{\mathrm{9}}{\mathrm{23}},\frac{\mathrm{16}}{\mathrm{23}}\right\} \\ $
${x}=\mathrm{27}−\frac{\mathrm{23}\left\{{x}\right\}−\mathrm{2}}{\mathrm{7}}+\left\{{x}\right\} \\ $
${x}\in\left\{\mathrm{27}\frac{\mathrm{2}}{\mathrm{23}},\mathrm{26}\frac{\mathrm{9}}{\mathrm{23}},\mathrm{25}\frac{\mathrm{16}}{\mathrm{23}}\right\} \\ $
Hope this acceptable...
