Find all values of $t$ such that $\lfloor t\rfloor = 3t + 4/5$. If you find more than one value, then list the values you find in increasing order, separated by commas.
The integers $\lfloor t\rfloor$ and $3t$ differ by at most 1, so $$|\lfloor t\rfloor - (3t + 4/5)| \le 1.$$Since 4/5 is less than 1, the right side of the above inequality is less than 2. We get $$|(3t + 4/5) - \lfloor t\rfloor| < 2.$$To make this absolute value less than 2, $3t$ must be at least $-2$ units away from the nearest integer. The possible values of $t$ are therefore as follows. Since $3t$ ranges from 0 to 3t+4/4 ranges from 0 plus 4/5 to plus 3 plus 4/5, so t ranges from $4/5$ to $19/5$. We now must intersect this interval with the interval $[-2,2]$. This intersection is $\boxed{[4/5,2]}$.
To find the values of $t$ that satisfy the equation $\lfloor t\rfloor = 3t + \frac{4}{5}$, we need to consider two cases:
1. When $t$ is an integer:
When $t$ is an integer, the value of $\lfloor t\rfloor$ is equal to $t$ itself. Therefore, we can rewrite the equation as $t = 3t + \frac{4}{5}$.
Now, solve for $t$:
$t - 3t = - \frac{4}{5}$
$-2t = - \frac{4}{5}$
$t = \frac{4}{5} \cdot \frac{1}{-2}$
$t = - \frac{2}{5}$
2. When $t$ is not an integer:
When $t$ is not an integer, we need to consider the floor function $\lfloor t \rfloor$ which rounds $t$ down to the nearest integer. In this case, $\lfloor t \rfloor$ is one less than the value of $t$.
Let's substitute this into the equation $\lfloor t \rfloor = 3t + \frac{4}{5}$:
$t - 1 = 3t + \frac{4}{5}$
Now, solve for $t$:
$4t = \frac{4}{5} - 1$
$4t = - \frac{1}{5}$
$t = - \frac{1}{5} \cdot \frac{1}{4}$
$t = - \frac{1}{20}$
So, the values of $t$ that satisfy the equation are $t = - \frac{2}{5}$ and $t = - \frac{1}{20}$.
In increasing order, the values are $t = - \frac{2}{5}, - \frac{1}{20}$.