The problem states:

• A rectangle has a length that is 10 less than 3 times the width.
• The area of the rectangle is 8 square feet.
• We are asked to find the width of the rectangle.

Let the width of the rectangle be $w$ feet. Then the length of the rectangle can be expressed as:

$\text{Length} = 3w - 10$

The area of a rectangle is given by:

$\text{Area} = \text{Length} \times \text{Width}$

Substitute the expressions for length and width into the area formula:

$8 = (3w - 10) \times w$

Now, solve for $w$:

$8 = 3w^2 - 10w$

Rearrange this into a standard quadratic form:

$3w^2 - 10w - 8 = 0$

Now, I will solve this quadratic equation.The solutions to the quadratic equation are $w = -\frac{2}{3}$ and $w = 4$.

Since the width of a rectangle cannot be negative, the width of the rectangle is $w = 4$ feet.

Would you like a detailed explanation or have any other questions?

Here are 5 related questions for practice:

1. How would you calculate the length of the rectangle given the width?
2. What if the area was doubled? How would that affect the width?
3. Can you form a similar problem where the length is a different multiple of the width?
4. What is the perimeter of the rectangle with the given dimensions?
5. How would this problem change if the area was in square meters instead of square feet?

Tip: When solving quadratic equations, always check for and discard non-physical solutions, like negative values when working with dimensions.