Let's break down the problem and solve it step-by-step.

Given Information:

• The area of the rectangle is $28 \, \text{yd}^2$.
• The length of the rectangle is 5 yards less than three times the width.

Step 1: Define variables

Let:

• $w$ represent the width of the rectangle (in yards).
• $l$ represent the length of the rectangle (in yards).

From the problem, we have:

• $l = 3w - 5$ (since the length is 5 yards less than three times the width).
• The area of the rectangle is given by $\text{Area} = l \times w$.

Substitute the expression for $l$ into the area formula: $\text{Area} = w \times (3w - 5)$ We are also given that the area is $28 \, \text{yd}^2$, so: $w \times (3w - 5) = 28$

Step 2: Solve the equation

Expand the left-hand side: $3w^2 - 5w = 28$ Rearrange the equation into standard quadratic form: $3w^2 - 5w - 28 = 0$

Step 3: Solve the quadratic equation

We can solve this using the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $3w^2 - 5w - 28 = 0$, we have $a = 3$, $b = -5$, and $c = -28$. Substituting these into the quadratic formula:

$w = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-28)}}{2(3)}$ $w = \frac{5 \pm \sqrt{25 + 336}}{6}$ $w = \frac{5 \pm \sqrt{361}}{6}$ $w = \frac{5 \pm 19}{6}$

Thus, the two possible solutions for $w$ are: $w = \frac{5 + 19}{6} = \frac{24}{6} = 4 \quad \text{(valid, positive width)}$ $w = \frac{5 - 19}{6} = \frac{-14}{6} \quad \text{(negative, not valid for width)}$

Step 4: Find the length

Now that we know $w = 4$, substitute this back into the expression for the length: $l = 3w - 5 = 3(4) - 5 = 12 - 5 = 7$

Final Answer:

• The width is $4 \, \text{yards}$.
• The length is $7 \, \text{yards}$.

Would you like further details or have any questions about this solution?

Here are some related questions to explore further:

1. How would the solution change if the area were different?
2. What if the length were 5 yards more than three times the width?
3. Can you solve the problem by factoring instead of using the quadratic formula?
4. How can we check the solution by calculating the area using the found dimensions?
5. How does the quadratic formula apply to different problems involving geometry?

Tip: When solving quadratic equations, always check whether the solutions make sense within the context of the problem (e.g., negative dimensions are not valid).