The problem describes a rectangle where the length is 3 yards less than twice the width, and the area of the rectangle is 14 square yards. We need to find the dimensions (length and width) of the rectangle.

Let:

• $w$ represent the width of the rectangle.
• $l$ represent the length of the rectangle.

From the problem, we are given two relationships:

1. The length is 3 yards less than twice the width, so: $l = 2w - 3$
2. The area of the rectangle is 14 square yards, and the area of a rectangle is given by $\text{Area} = l \times w$. Therefore: $l \times w = 14$

Substitute the expression for $l$ from the first equation into the second equation: $(2w - 3) \times w = 14$ Expand and solve for $w$: $2w^2 - 3w = 14$ $2w^2 - 3w - 14 = 0$

This is a quadratic equation. We can solve it using the quadratic formula: $w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-14)}}{2(2)}$ Simplify: $w = \frac{3 \pm \sqrt{9 + 112}}{4}$ $w = \frac{3 \pm \sqrt{121}}{4}$ $w = \frac{3 \pm 11}{4}$

So, $w$ can be: $w = \frac{3 + 11}{4} = \frac{14}{4} = 3.5$ or $w = \frac{3 - 11}{4} = \frac{-8}{4} = -2$

Since the width cannot be negative, $w = 3.5$ yards.

Now, using the first equation to find the length: $l = 2w - 3 = 2(3.5) - 3 = 7 - 3 = 4 \text{ yards}.$

Therefore, the dimensions of the rectangle are:

• Width: $3.5$ yards
• Length: $4$ yards

Would you like more details, or do you have any questions?

Here are five related questions to explore:

1. How would you check the solution by calculating the area again?
2. What would happen if the length was 3 yards more than twice the width?
3. Can you solve the quadratic equation using factoring instead of the quadratic formula?
4. How would the solution change if the area was different but the length relation stayed the same?
5. What are other possible ways to describe the dimensions of the rectangle?

Tip: Always check for possible extraneous solutions, especially when solving quadratic equations, as some results might not be valid for the context (e.g., negative dimensions).