Question

the length of a rectangle is 3 yd less than twice the width, and the area of the rectangle is 65 yd². find the dimensions of the rectangle. length : yd width : yd

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ yd. Then the length $l = 2w - 3$ yd.

Step2: Set up the area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 65$ yd², we substitute $l$ and $A$ into the formula: $(2w - 3)\times w=65$.
Expand the left - hand side: $2w^{2}-3w = 65$.
Rearrange to get a quadratic equation: $2w^{2}-3w - 65 = 0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 2$, $b=-3$, $c = - 65$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor.
Factor the quadratic equation: $2w^{2}-3w - 65=(2w + 13)(w - 5)=0$.
Set each factor equal to zero:
$2w+13 = 0$ gives $w=-\frac{13}{2}$, but the width cannot be negative, so we discard this solution.
$w - 5=0$ gives $w = 5$ yd.

Step4: Find the length

Substitute $w = 5$ into the length formula $l=2w - 3$.
$l=2\times5-3=10 - 3=7$ yd.

Answer:

Length: 7 yd
Width: 5 yd