# Solving Equations Containing Floor Functions

Recently I have been struggling with a problem involving the floor function. The problem is: $$\lfloor x+5 \rfloor = 3\lfloor x\rfloor-1$$

I have had a similar question to this however it only involved the floor function on one side of the equation and I was able to set up a pair of inequalities and solve them however I have not been able to do the same for this question.

What I have done so far is attempt the same procedure by solving the equation and then setting up a set of inequalities but have been unable to come up with a reasonable answer. I have also graphed this for clarity but am still unsure what is really going on or how to approach this question.

Thank you.

• It's harder when the multiplication is inside the floor. – marty cohen Aug 25 '17 at 3:55

This is equivalent to $\lfloor x\rfloor +5=3\lfloor x\rfloor-1$, i.e. $$\lfloor x\rfloor=3 \implies x \in [3,4[.$$
• equation: $$\lfloor x+5 \rfloor=3\lfloor x \rfloor−1$$
• same as: $$\lfloor x\rfloor+5 = 3\lfloor x \rfloor-1$$
• add $1$: $$\lfloor x \rfloor+6 = 3\lfloor x \rfloor$$
• subtract $\lfloor x \rfloor$: $$6 = 2\lfloor x \rfloor$$
• divide by $2$: $$3 = \lfloor x \rfloor$$
• do the floor thing: $$3 \le x \lt 4$$