Solve algebraically using only variable: The length of a rectangle is two less than three times

its width. If the area of the rectangle is 65, find the length and the width. (The area of a rectangle is equal to its length times its width.

w(3w-2) = 65

5*13 = 65

To solve this problem algebraically, we can set up equations using the given information.

Let's represent the width of the rectangle with the variable "w."
According to the problem, the length of the rectangle is two less than three times its width. Therefore, the length can be represented as "3w - 2."

The area of a rectangle is equal to its length times its width. So, we can set up the equation:

Area = Length × Width

Using the values from the problem, we have:

65 = (3w - 2) × w

Now, we can simplify the equation and solve for "w."

65 = 3w^2 - 2w

To solve a quadratic equation, we need to set it equal to zero:

3w^2 - 2w - 65 = 0

At this point, we can use factoring, completing the square, or the quadratic formula to find the value(s) of "w." In this case, let's use the factoring method.

The equation can be factored as:

(3w + 13)(w - 5) = 0

Now, we set each factor equal to zero separately:

3w + 13 = 0 or w - 5 = 0

Solving these equations will give us the possible values of "w."

For 3w + 13 = 0: subtracting 13 from both sides, we get 3w = -13, then dividing by 3, we have w = -13/3.

For w - 5 = 0: adding 5 to both sides, we get w = 5.

So, we have two possible values for the width: w = -13/3 or w = 5.

Since the width of a rectangle cannot be negative, we discard w = -13/3.

Therefore, the width of the rectangle is w = 5.

To find the length, we substitute the value of w back into the length equation:

Length = 3w - 2

Length = 3(5) - 2

Length = 15 - 2

Length = 13

Thus, the width of the rectangle is 5 units, and the length is 13 units.