To solve this problem, let's analyze the equations and the geometry of the rectangle:

Problem Breakdown:

1. Given:

   • ABCD is a rectangle.
   • AC (one diagonal) is expressed as $\frac{3y}{4}$.
   • BD (the other diagonal) is expressed as $6y - 3$.

2. Key Property of a Rectangle:

   • In a rectangle, the diagonals are equal in length. Hence, $AC = BD$.

3. Set the equations equal: $\frac{3y}{4} = 6y - 3$

Solution:

Step 1: Eliminate the fraction by multiplying through by 4:

$3y = 4(6y - 3)$

Step 2: Simplify:

$3y = 24y - 12$

Step 3: Rearrange to isolate $y$:

$3y - 24y = -12$ $-21y = -12$

Step 4: Solve for $y$:

$y = \frac{-12}{-21} = \frac{4}{7}$

Step 5: Calculate the lengths of AC and BD:

• Substitute $y = \frac{4}{7}$ into the expressions for $AC$ and $BD$:
  • $AC = \frac{3y}{4} = \frac{3 \cdot \frac{4}{7}}{4} = \frac{12}{28} = \frac{3}{7}$.
  • $BD = 6y - 3 = 6 \cdot \frac{4}{7} - 3 = \frac{24}{7} - 3 = \frac{24}{7} - \frac{21}{7} = \frac{3}{7}$.

Thus, the length of each diagonal is $\boxed{\frac{3}{7}}$.

Would you like a detailed step-by-step explanation of any part? Let me know if you have questions or need clarification!

Related Questions:

1. How can we use the property of diagonals to determine unknown variables in other quadrilaterals?
2. If one diagonal's formula includes a square term, how would the solution process change?
3. What are some real-life applications of rectangle diagonal properties?
4. How would the problem change if ABCD were a square instead of a rectangle?
5. Can you solve for $y$ if one diagonal was twice the other?

Tip: Always check the geometry properties of the shape when solving problems involving diagonals, sides, or angles.