The problem involves a square and an equilateral triangle, both with the same perimeter. Each side of the square is labeled $2x + 5$, and each side of the equilateral triangle is labeled $2x + 8$.

Step 1: Set up the equations for the perimeters.

• The perimeter of a square is calculated as $4 \times \text{side length}$.

  • For the square, the perimeter is $4(2x + 5)$.

• The perimeter of an equilateral triangle is calculated as $3 \times \text{side length}$.

  • For the equilateral triangle, the perimeter is $3(2x + 8)$.

Step 2: Set up the equation because the perimeters are equal.

$4(2x + 5) = 3(2x + 8)$

Step 3: Expand both sides.

• Left side: $4(2x + 5) = 8x + 20$
• Right side: $3(2x + 8) = 6x + 24$

So the equation becomes: $8x + 20 = 6x + 24$

Step 4: Solve for $x$.

Subtract $6x$ from both sides: $8x - 6x + 20 = 24$ $2x + 20 = 24$ Subtract 20 from both sides: $2x = 4$ Divide by 2: $x = 2$

Step 5: Calculate the perimeter of each shape.

• For the square, the side length is $2x + 5 = 2(2) + 5 = 4 + 5 = 9$. Therefore, the perimeter is $4 \times 9 = 36$.
• For the triangle, the side length is $2x + 8 = 2(2) + 8 = 4 + 8 = 12$. Therefore, the perimeter is $3 \times 12 = 36$.

Final Answer:

• The value of $x$ is $2$.
• The perimeter of each shape is $36$.

Would you like further details or clarification on any part of this solution? Here are 5 related questions:

1. How do you calculate the perimeter of polygons like squares and triangles?
2. What are the steps to solve a system of linear equations?
3. Can the equation setup be different for irregular polygons?
4. How does the relationship between side lengths and perimeters apply to other shapes?
5. What happens if one of the shapes was a rectangle or an isosceles triangle?

Tip: When solving word problems, always translate the relationships into equations before simplifying and solving.