The problem you've shared involves comparing the areas of a rectangle and a triangle to solve for $x$. Let's break it down step by step:

• The rectangle has:

  • Length = $(x + 1)$ feet
  • Width = 5 feet

  The area of the rectangle is: $\text{Area of rectangle} = \text{Length} \times \text{Width} = (x + 1) \times 5 = 5(x + 1) \, \text{square feet}$

• The triangle has:

  • Base = 4 feet
  • Height = $(x + 10)$ feet

  The area of the triangle is: $\text{Area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 4 \times (x + 10) = 2(x + 10) \, \text{square feet}$

According to the problem, the area of the rectangle is equal to the area of the triangle. So, we set the two areas equal to each other and solve for $x$:

$5(x + 1) = 2(x + 10)$

Now, let's solve this equation.

$5(x + 1) = 2(x + 10)$ Expanding both sides:

$5x + 5 = 2x + 20$

Next, subtract $2x$ from both sides:

$3x + 5 = 20$

Now, subtract 5 from both sides:

$3x = 15$

Finally, divide by 3:

$x = 5$

Thus, the value of $x$ is 5.

Would you like more details or further clarifications on this? Here are some related questions to expand the concept:

1. How do you calculate the area of different polygons?
2. How does changing the base or height of a triangle affect its area?
3. What other methods can we use to solve linear equations like this?
4. How would the solution change if the dimensions of the triangle or rectangle were altered?
5. How can this concept be applied to real-world problems involving area?

Tip: Always double-check your algebraic manipulations, especially when distributing and combining like terms, to avoid small errors.