#### Answer

$BD = 1$
$AC = 1$

#### Work Step by Step

According to Theorem 6-15, the diagonals of a rectangle are congruent. Therefore, we can set $AC$ and $BD$, the diagonals of $ABCD$, equal to one another to solve for $y$:
$AC = BD$
$\frac{3y}{5} = 3y - 4$
Multiply both sides of the equation by $5$ to get rid of the fraction:
$3y = 15y - 20$
Subtract $15y$ from each side of the equation to move the variable to the left side of the equation:
$-12y = -20$
Divide each side of the equation by $-12$:
$y = \frac{20}{12}$
$y = \frac{5}{3}$
Now that we have the value of $y$, we can plug $\frac{5}{3}$ in for $y$. We do not need to evaluate both expressions because we know that the lengths of the two diagonals are equal to one another:
$BD = 3(\frac{5}{3}) - 4$
Cancel the $3$ to simplify the fraction:
$BD = 5 - 4$
Subtract to solve:
$BD = 1$
If $BD = 1$, then $AC = 1$ because diagonals of rectangles are congruent.