# Floor function algebra 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?

• You can get the floor functions with \lfloor and \rfloor, e.g., \lfloor \pi \rfloor = 3. If you need bigger ones, as around a fraction, for instance, use \left\lfloor and \right\rfloor, e.g., \left\lfloor\frac{4x+7}{3}\right\rfloor. May 21 '20 at 3:24
• You seem to be assuming that $4x+7$ is an integer. There's no reason to think it is. May 21 '20 at 3:37
• X must be of the form $$10{\alpha}+6;\alpha \inf \mathbb{Z}$$ May 21 '20 at 4:53
• @saulspatz so would it be safe to say that the RHS is in the range of $(4x+7)/3$ to $(4x+7)/3-0.99$ ?
– user377742
May 21 '20 at 8:52
• Look at my answer. May 21 '20 at 16:09

I'll get you started.

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

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.

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

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 $$x$$.

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.