We are given the following information:

1. Rectangle Dimensions:

   • Length = $5 + 2x$ inches
   • Width = 10 inches

2. Triangle Dimensions:

   • Base = 30 inches
   • Height = $4x - 10$ inches

Step 1: Write the expressions for the areas

• The area $A_{\text{rectangle}}$ of a rectangle is given by: $A_{\text{rectangle}} = \text{Length} \times \text{Width} = (5 + 2x) \times 10$ So, $A_{\text{rectangle}} = 10(5 + 2x) = 50 + 20x$

• The area $A_{\text{triangle}}$ of a triangle is given by: $A_{\text{triangle}} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 30 \times (4x - 10)$ Simplifying, $A_{\text{triangle}} = 15(4x - 10) = 60x - 150$

Step 2: Set the areas equal

Since the area of the rectangle is equal to the area of the triangle, we can set the two expressions equal to each other: $50 + 20x = 60x - 150$

Step 3: Solve for $x$

1. Subtract $20x$ from both sides: $50 = 40x - 150$

2. Add 150 to both sides: $200 = 40x$

3. Divide by 40: $x = 5$

Step 4: Check the result

• For $x = 5$:

  • Rectangle: $5 + 2(5) = 5 + 10 = 15$ inches (length).

  • Area of the rectangle = $15 \times 10 = 150$ square inches.

  • Triangle: $4(5) - 10 = 20 - 10 = 10$ inches (height).

  • Area of the triangle = $\frac{1}{2} \times 30 \times 10 = 150$ square inches.

Since both areas are 150 square inches, the solution is correct.

Final Answer: $x = 5$.

Would you like further clarification on any of the steps?

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Here are 5 related questions you might consider:

1. How do we derive the formula for the area of a rectangle?
2. Can we solve for the dimensions of the triangle using the found value of $x$?
3. What would happen if the rectangle and triangle had different areas?
4. How can we solve similar problems with other geometric shapes?
5. How can we express the area of a rectangle and triangle using algebraic terms?

Tip: Always remember to verify your solution by substituting the values back into the original problem.