A rectangular garden has a perimeter of 56 ft. The length of the garden is one more than twice the width. What are the dimensions of the garden?
Let's say the width of the garden is x feet.
So, the length of the garden is (2x + 1) feet.
The perimeter of a rectangle is given by the formula: P = 2(length + width).
Substituting the given values, we have: 56 = 2((2x + 1) + x).
Simplifying the equation, we get: 56 = 2(3x + 1).
Expanding the product on the right side, we have: 56 = 6x + 2.
Subtracting 2 from both sides of the equation, we get: 54 = 6x.
Dividing both sides of the equation by 6, we get: x = 9.
So, the width of the garden is 9 feet, and the length is (2 * 9 + 1) = 19 feet. Answer: \boxed{9 \text{ feet}, 19 \text{ feet}}.
To find the dimensions of the garden, let's set up equations based on the given information:
Let L be the length of the garden and W be the width of the garden.
According to the problem, the perimeter of the garden is 56 ft, which can be written as:
2L + 2W = 56 --- [Equation 1]
The problem also states that the length of the garden is one more than twice the width, which can be written as:
L = 2W + 1 --- [Equation 2]
Now we have a system of two equations (Equation 1 and Equation 2). We can solve this system to find the values of L and W.
First, substitute Equation 2 into Equation 1:
2(2W + 1) + 2W = 56
Distribute:
4W + 2 + 2W = 56
Combine like terms:
6W + 2 = 56
Subtract 2 from both sides:
6W = 54
Divide both sides by 6:
W = 9
Now, substitute the value of W back into Equation 2 to find the value of L:
L = 2W + 1 = 2(9) + 1 = 18 + 1 = 19
Therefore, the dimensions of the garden are 19 ft (length) and 9 ft (width).