## Geometry: Common Core (15th Edition)

$BD = 1$ $AC = 1$
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.