Let's solve this problem by analyzing the given shapes and their areas.

Step 1: Understanding the Shapes and Their Dimensions

• Rectangle:

  • Length: $y + 3$
  • Width: $y$
  • Area of the rectangle = Length × Width = $y(y + 3)$.

• Triangle:

  • Base: $2y + 1$
  • Height: $y + 1$
  • Area of the triangle = $\frac{1}{2} \times \text{Base} \times \text{Height}$ $\text{Area of triangle} = \frac{1}{2} \times (2y + 1)(y + 1)$

Step 2: Set Up the Equation

The problem states that the area of the rectangle is equal to the area of the triangle: $y(y + 3) = \frac{1}{2} \times (2y + 1)(y + 1)$

Step 3: Simplify the Equation

Multiply both sides by 2 to eliminate the fraction: $2y(y + 3) = (2y + 1)(y + 1)$

Expand both sides:

• Left side: $2y^2 + 6y$.
• Right side: Use the distributive property: $(2y + 1)(y + 1) = 2y^2 + 2y + y + 1 = 2y^2 + 3y + 1$

So, the equation becomes: $2y^2 + 6y = 2y^2 + 3y + 1$

Step 4: Solve for $y$

Subtract $2y^2$ from both sides: $6y = 3y + 1$

Subtract $3y$ from both sides: $3y = 1$

Divide by 3: $y = \frac{1}{3}$

Step 5: Verify the Solution

Substitute $y = \frac{1}{3}$ back into the dimensions to ensure that both areas are equal.

Would you like a detailed verification or have any other questions?

Related Questions:

1. What would happen if $y$ was negative in this problem?
2. How do you find the area of other geometric shapes like circles or trapezoids?
3. How can we factor quadratic equations like the one in this problem?
4. What are other scenarios where the area of different shapes might be equal?
5. How would this problem change if the triangle had a different formula for the area?

Tip:

When dealing with equations involving fractions, try multiplying both sides by the least common multiple to simplify calculations.