The length of a rectangle is 6 inches less than 3 times its width. Find the dimensions of the rectangle if its area is 45 square inches.

W = X in.

L = (3x - 6) in.

A = L*W = (3x-6)X = 45,
3x^2 - 6x = 45
3x(x-2) = 45,
x(x-2) = 15,
x^2 - 2x - 15 = 0
(x-5)(x+3) = 0,
x-5 = 0,
x = 5 in.

x+3 = 0,
x = -3.

Solution Set: x = 5, and x = -3.

Select + value of X:
x = 5

W = x = 5 in.
L = 3x-6 = 3*5 - 6 = 9 in.

To find the dimensions of the rectangle, we can set up an equation based on the given information. Let's denote the width of the rectangle as "w" and the length as "l."

According to the problem, the length of the rectangle is 6 inches less than 3 times its width, which can be represented by the equation:

l = 3w - 6 --(1)

We also know that the area of a rectangle is given by the formula:

Area = length * width

Substituting the values of length (l) and width (w) in the equation, we have:

45 = (3w - 6) * w

Now, let's solve this equation for the width of the rectangle.

45 = (3w - 6) * w
45 = 3w^2 - 6w

Rearranging the equation, we get a quadratic equation in standard form:

3w^2 - 6w - 45 = 0

To solve this quadratic equation, we can factor it, complete the square, or use the quadratic formula. In this case, let's solve it by factoring.

Factoring the quadratic equation, we have:

(3w + 9)(w - 5) = 0

Setting each factor equal to zero, we get two possible values for the width:

3w + 9 = 0 or w - 5 = 0

Solving each equation, we find:

3w = -9 or w = 5

Dividing both sides of the first equation by 3, we have:

w = -3

However, since we're dealing with dimensions, the width cannot be negative. Therefore, we discard the negative value of the width (w = -3).

Thus, the width of the rectangle is 5 inches.

Plugging this value of the width into equation (1) to find the length (l), we get:

l = 3w - 6
l = 3(5) - 6
l = 15 - 6
l = 9

Therefore, the dimensions of the rectangle are: width = 5 inches and length = 9 inches.