1. The length of a rectangle is twice its width. The perimeter of the rectangle is 24 inches. Write a system of equations in 2 variables. Use substitution to solve the problem.

A. Equation 1:

Equation 2:
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B. What is the area of the rectangle?

A. Equation 1:

P = 2L + 2W
24 = 2(2W) + 2W
24 = 6W
4 = W
? = L

A = LW

A. Let's denote the width of the rectangle as 'w' (in inches). Since the length is twice the width, we can write the length as '2w' (in inches).

The perimeter of a rectangle is given by the formula: 2(length + width). In this case, the perimeter is given as 24 inches. So, we can set up the first equation:

Equation 1:
2(2w + w) = 24

Simplifying Equation 1, we have:
6w = 24

Now, we can solve for 'w' to find the width.

Next, we can substitute the value of 'w' back into the equation to find the length.

Equation 2:
Length = 2w

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B. To find the area of the rectangle, we use the formula: A = length * width. Since we have already found the values of length and width in the previous step, we can substitute them into the formula to find the area.