the length of a rectangle is 1 inch less than twice the width, and the perimeter is 40 inches.find the width and the length.

Ah, the classic tale of the mysterious rectangle with its algebraic secrets! Now, let me put on my detective hat and solve this case for you.

Let's call the width of the rectangle "W." According to our given information, the length is one inch less than twice the width, meaning the length is (2W - 1).

Now, if we remember our handy-dandy formula for the perimeter of a rectangle, it's 2 times the width plus 2 times the length. Plugging in our values, we get:

Perimeter = 2W + 2(2W - 1)

But wait, there's more! We're also told that the perimeter of our rectangle is 40 inches. So, let's solve for the width:

40 = 2W + 2(2W - 1)

Now, let me consult the "Guide to Solving Equations" issued by the Society of Silly Mathematicians. It tells me that, after performing some arithmetic magic, the width of our rectangle is 7 inches.

Once we know the width, we can easily find the length by substituting the value of W back into our equation: Length = 2W - 1. In this case, the length would be 2(7) - 1 = 13 inches.

So, my dear friend, the width of the rectangle is 7 inches, and the length is 13 inches. Mystery solved!

Let's denote the width of the rectangle as "w" and the length as "l".

According to the given information, we have two equations:

1) l = 2w - 1 (the length is 1 inch less than twice the width)
2) 2(l + w) = 40 (the perimeter is 40 inches)

We can substitute equation 1 into equation 2 to solve for w:

2((2w - 1) + w) = 40
2(3w - 1) = 40
6w - 2 = 40
6w = 42
w = 7

Now, we can substitute the value of w back into equation 1 to find the length:

l = 2w - 1
l = 2(7) - 1
l = 14 - 1
l = 13

Therefore, the width of the rectangle is 7 inches and the length is 13 inches.

To solve this problem, let's first assign variables to the width and length of the rectangle.

Let's say that the width of the rectangle is "w" inches, and the length of the rectangle is "l" inches.

According to the problem:

1. The length of the rectangle is 1 inch less than twice the width.
This can be expressed as l = 2w - 1.

2. The perimeter of a rectangle is given by the formula: P = 2(l + w), where P represents the perimeter.
In this case, the perimeter is given as 40 inches. So we can write the equation as 40 = 2(l + w).

Now we have a system of two equations with two variables. We can solve them simultaneously to find the values of "w" and "l".

1. Substitute the value of "l" from equation 1 in terms of "w" into equation 2:
40 = 2((2w - 1) + w)

2. Simplify the equation:
40 = 2(3w - 1)
40 = 6w - 2

3. Solve for "w":
6w = 42 (added 2 to both sides)
w = 7 (divided both sides by 6)

4. Substitute the value of "w" back into equation 1 to find "l":
l = 2w - 1 (substituting w = 7)
l = 2(7) - 1
l = 14 - 1
l = 13

Therefore, the width of the rectangle is 7 inches, and the length is 13 inches.

let width be x

then length is 2x-1

so isn't 2(x) + 2(2x-1) = 40 ?
solve for x