Floor function algebra question

Question

I don't know how to put floor functions in but...

Solve $$\dfrac{19x + 16}{10} = \left \lfloor \dfrac{4x+7}{3}\right \rfloor$$

I have so far worked out that the RHS can either be $(4x+7)/3 - 0.33$, $(4x+7)/3 - 0.67$ or itself. When I solve for each of these three equations, I get $x=12/17, 22/17, 2/17$. From there, I subbed $x$ back into the equation to try and see which one works but none did. Can I have some help?

Answer 1

I'll get you started.

We know that $x-1<\lfloor x\rfloor<=x$, so we have $$ \frac{4x+4}3<\frac{19x+16}{10}\leq\frac{4x+7}3\\ 40x+40<57x+48\leq40x+70\\ \frac{-8}{17}<x\leq\frac{22}{17}$$ so that $x=n+\varepsilon$ where $n\in\{-1,0,1\}$ and $0\leq\varepsilon<1$.

Now we can test each of the three possibilities for $n$ separately. Suppose $x=1+\varepsilon$. Then $$\frac{19x+16}{10}=\frac{35+19\varepsilon}{10}$$ is an integer between $3.5$ and $5.4$ so there are only two possibilities for $\varepsilon$. Check these to see if $x=1+\varepsilon$ satisfies the equation. Repeat the process for $n=0$ and $n=-1$.

Answer 2

Since algebra with floor functions is rarely nice, I like to graph it if possible to visualize the solutions. In your case, this is the graph of $$\frac{19x+16}{10}-\Big\lfloor \frac{4x+7}{3}\Big\rfloor$$

(This is using Desmos by the way) Notice how any solutions would pass through the $y=0$, and there seem to be $4$. One seems to occur at $x \approx -.3158$ and by plugging this into the floor function we see it approaches $1.912$, the floor of which is $1$. So we are looking for $x$ such that $\frac{19x+16}{10} = 1$. The solution is $x = \frac{-6}{19}$ which, by substituting it back in, works. Similarly, we can show that we need $x \approx .2105$ such that $\frac{19x+16}{10} = 2$, giving the solution $x = \frac{4}{19}$. Then we need $x \approx .7368$ such that $\frac{19x+16}{10} = 3$, giving the solution $x = \frac{14}{19}$. Finally, we need $x \approx 1.2632$ such that $\frac{19x+16}{10} = 4$, giving the solution $x = \frac{24}{19}$. From here, you can use strict inequalities to prove that there are no more solutions.

Answer 3

$$\dfrac{19x + 16}{10} = \left \lfloor \dfrac{4x+7}{3}\right \rfloor$$

Let $\dfrac{19x + 16}{10} = n \in \mathbb Z$.

Then $x = \dfrac{10n-16}{19}$

and $\dfrac{4x+7}{3} = \dfrac{40n+69}{57}$.

So

$$n \le \dfrac{40n+69}{57} < n + 1$$

$$57n \le 40n+69 < 57n + 57$$

$$ 0 \le -17n+69 < 57$$

$$ -69 \le -17n < -12$$

$$ \dfrac{12}{17} < n \le 4\dfrac{1}{17}$$

So now you can find the values of $n$ and then the values of $x$.

Answer 4

First deal with the fact that $\dfrac{19x + 16}{10}$ is an integer $k$.

So $19x = 10k -16$ is and integer $m$ and $x = \frac m{19}$ for some integer $m$.

Now deal with $\dfrac{19x + 16}{10}=\lfloor \dfrac{4x+7}{3} \rfloor$ so

$\dfrac{19x + 16}{10} \le \dfrac{4x+7}{3}< \dfrac{19x + 16}{10}+1=\dfrac{19x +26}{10}$

Replace $x$ with $\frac m{19}$ and

$\dfrac {m + 16}{10} \le \dfrac {\frac 4{19}m + 7}3 < \frac {m+26}{10}$

$3m + 48 \le \frac 40{19}m +70 < 3m + 78$

$-22 \le \frac {40}{19}m - 3m < 8$

$-22*19 \le 40m - 57m = -17*m < 8*19$

$\frac {-8*19}{17} < m \le \frac {22*19}{17}$

$-8 \frac {16}{17} < m \le 24 \frac {10}{17}$

So $-8 < m \le 24$

But $m = 10k-16$ so $m\equiv -16\equiv -6 \equiv 4 \pmod{10}$ so so

$m =-6,4,14,24$ are acceptable values.

And $x =-\frac 6{19},\frac 4{19}, \frac {14}{19}$ and $\frac {24}{29}$ are acceptable answers.

Answer 5

$$\lfloor\frac{\mathrm{4}{x}+\mathrm{7}}{\mathrm{3}}\rfloor=\frac{\mathrm{19}{x}+\mathrm{16}}{\mathrm{10}}\in{Z} \\ $$ $$\frac{\mathrm{4}{x}+\mathrm{7}}{\mathrm{3}}−\frac{\mathrm{19}{x}+\mathrm{16}}{\mathrm{10}}=\left[\mathrm{0};\mathrm{1}\right) \\ $$ $$\frac{\mathrm{22}−\mathrm{17}{x}}{\mathrm{30}}=\left[\mathrm{0};\mathrm{1}\right) \\ $$ $${x}=\frac{\mathrm{22}−\mathrm{30}\left[\mathrm{0};\mathrm{1}\right)}{\mathrm{17}}=\frac{\left(-\mathrm{8};\mathrm{22}\right]}{\mathrm{17}} \\ $$ $$\frac{\frac{\mathrm{19}}{\mathrm{17}}\left(-\mathrm{8};\mathrm{22}\right]+\mathrm{16}}{\mathrm{10}}=\left[\mathrm{1};\mathrm{4}\right] \\ $$ $$\frac{\mathrm{19}{x}+\mathrm{16}}{\mathrm{10}}=\left[\mathrm{1};\mathrm{4}\right] \\ $$ $${x}=\frac{\mathrm{10}\left[\mathrm{1};\mathrm{4}\right]−\mathrm{16}}{\mathrm{19}} \\ $$ $${x}=\left\{\frac{-\mathrm{6}}{\mathrm{19}},\frac{\mathrm{4}}{\mathrm{19}},\frac{\mathrm{14}}{\mathrm{19}},\frac{\mathrm{24}}{\mathrm{19}}\right\} \\ $$