A rectangle has a perimeter of 24 and an area of 35 square feet. What is the length and width of the rectangle?

length --- x

width ---- y

2x + 2y = 24
x + y = 12
y = 12-x

xy=35
x(12-x) = 35
-x^2 + 12x - 35 = 0
x^2 - 12x + 35 = 0
(x-7)(x-5)= 0
x = 7 or x = 5

if x=7, y=5 or if x=5, then y=7

the rectangle is 5 by 7

Well, it seems like this rectangle has some quirky dimensions! To find the length and width, let's use a bit of clown math, shall we?

Let's call the length "L" and the width "W". Now we know that the perimeter of the rectangle is 24. So, using some clown tricks, we can say:

2L + 2W = 24

Next, we know that the area of the rectangle is 35 square feet. Using a little clown magic, we have:

L * W = 35

Now, let's put on our clown noses and solve these equations! By manipulating the first equation, we can rearrange it to express L in terms of W:

L = 12 - W

Now we can substitute this expression into the second equation:

(12 - W) * W = 35

Let's simplify:

12W - W^2 = 35

Rearranging again:

W^2 - 12W + 35 = 0

Now, by using some clown algebra, we can factorize this equation as:

(W - 5)(W - 7) = 0

So we have two possible solutions: W could be either 5 or 7.

If W is 5, then the length L would be 12 - 5 = 7. And if W is 7, the length L would be 12 - 7 = 5.

So, the possible dimensions of the rectangle are 5 feet by 7 feet or 7 feet by 5 feet. Time to bring in the clown carpenter to build this quirky rectangle!

To find the length and width of the rectangle, we can use the formulas for perimeter and area.

Let's assume the length of the rectangle is "l" and the width is "w".

The formula for perimeter is: P = 2l + 2w
Given that the perimeter is 24, we can write the equation as: 24 = 2l + 2w

The formula for area is: A = l * w
Given that the area is 35, we can write the equation as: 35 = l * w

Now we have a system of two equations:
1) 24 = 2l + 2w
2) 35 = l * w

To solve this system of equations, we can use substitution or elimination method.

Let's solve the first equation for l:
2l = 24 - 2w
l = 12 - w/2

Now substitute this value of l in the second equation:
35 = (12 - w/2) * w
35 = 12w - w^2/2

Rearrange the equation:
w^2/2 - 12w + 35 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring may not be possible here, so let's use the quadratic formula:
w = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1/2, b = -12, c = 35

Substituting the values:
w = (-(-12) ± sqrt((-12)^2 - 4 * (1/2) * 35)) / 2 * (1/2)
w = (12 ± sqrt(144 - 70)) / 1
w = (12 ± sqrt(74)) / 1

Now, we have two possible values for the width: w = 12 + sqrt(74) or w = 12 - sqrt(74).

To find the corresponding length, substitute the values of w back into the equation for perimeter:

For the width w = 12 + sqrt(74):
l = 12 - (12 + sqrt(74)) / 2
l = 12 - 6 - sqrt(74) / 2
l = 6 - sqrt(74) / 2

For the width w = 12 - sqrt(74):
l = 12 - (12 - sqrt(74)) / 2
l = 12 - 6 + sqrt(74) / 2
l = 6 + sqrt(74) / 2

Therefore, the possible length and width combinations are:
1) Length = 6 - sqrt(74)/2 and Width = 12 + sqrt(74)
2) Length = 6 + sqrt(74)/2 and Width = 12 - sqrt(74)

These are the step-by-step instructions to find the length and width of the given rectangle.

To find the length and width of the rectangle, we can use the formulas for perimeter and area.

The perimeter of a rectangle is given by the formula:
perimeter = 2(length + width)

The area of a rectangle is given by the formula:
area = length * width

In this case, we are given that the perimeter is 24 and the area is 35 square feet.

Using the perimeter formula, we can write:
24 = 2(length + width)

Using the area formula, we can write:
35 = length * width

We now have a system of two equations:
24 = 2(length + width)
35 = length * width

To solve this system of equations, we can use substitution or elimination.

Let's solve it using substitution method:

From the first equation, we can express one variable in terms of the other variable. For example, we can solve for length in terms of width:

24 = 2(length + width)
12 = length + width
12 - width = length

Now we can substitute this expression for length into the second equation:

35 = length * width
35 = (12 - width) * width

Expanding and simplifying:
35 = 12w - w^2

Rearranging the equation:
w^2 - 12w + 35 = 0

This is a quadratic equation. We can now solve it by factoring or using the quadratic formula.
Factoring:
(w - 7)(w - 5) = 0

Setting each factor equal to zero gives us two possible values for the width:

w - 7 = 0 or w - 5 = 0

Solving for w, we find:
w = 7 or w = 5

Now we can find the corresponding length values using the expression we found earlier:

length = 12 - width

For w = 7:
length = 12 - 7 = 5

For w = 5:
length = 12 - 5 = 7

So, the possible length and width combinations for the rectangle are:
Length = 5 ft, Width = 7 ft
or
Length = 7 ft, Width = 5 ft

A = lw

35 = 5 * 7