a. Writing the system of equations

Let:

• $L$ be the length of the rectangle.
• $W$ be the width of the rectangle.

From the problem, we know the following:

1. The length of the rectangle is twice the width:
   $L = 2W \quad \text{(Equation 1)}$

2. The perimeter of the rectangle is 24 inches. The formula for the perimeter of a rectangle is:
   $P = 2L + 2W$ Substituting the given perimeter: $2L + 2W = 24 \quad \text{(Equation 2)}$

b. Solving the system of equations

Step 1: Substitution

• From Equation 1, $L = 2W$. Substitute this expression for $L$ in Equation 2:

$2(2W) + 2W = 24$

Step 2: Simplify and solve for $W$: $4W + 2W = 24$ $6W = 24$ $W = \frac{24}{6} = 4$

Step 3: Find $L$: Now that $W = 4$, substitute this into Equation 1 to find $L$: $L = 2W = 2(4) = 8$

c. Finding the area of the rectangle

The area $A$ of a rectangle is given by: $A = L \times W$ Substitute $L = 8$ and $W = 4$: $A = 8 \times 4 = 32 \, \text{square inches}$

Final Answer:

The area of the rectangle is 32 square inches.

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Would you like more details or have any questions about this solution?

Here are 5 follow-up questions:

1. How do you derive the formula for the perimeter of a rectangle?
2. Can this method be applied to solve for other shapes with different perimeter conditions?
3. What happens if the perimeter changes to 30 inches instead of 24?
4. How would the system of equations change if the length was three times the width?
5. Can you graph the equations to visualize the solution?

Tip: In substitution problems, always start by solving one equation for a single variable, then substitute it into the other equation to reduce the system to one variable.