Question

the length of a rectangle is 7 yd less than three times the width, and the area of the rectangle is 66 yd². find the dimensions of the rectangle.
length: yd
width: yd

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ yards. Then the length $l = 3w - 7$ yards.

Step2: Set up area - formula equation

The area of a rectangle $A=l\times w$. Given $A = 66$ yd², we substitute $l$ and $A$ into the formula: $(3w - 7)\times w=66$.

Step3: Expand the equation

$3w^{2}-7w = 66$, which can be rewritten as $3w^{2}-7w - 66=0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 3$, $b=-7$, $c = - 66$), we use the quadratic formula $w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-7)^{2}-4\times3\times(-66)=49 + 792=841$. Then $w=\frac{7\pm\sqrt{841}}{2\times3}=\frac{7\pm29}{6}$.
We get two solutions for $w$: $w_1=\frac{7 + 29}{6}=\frac{36}{6}=6$ and $w_2=\frac{7-29}{6}=\frac{-22}{6}=-\frac{11}{3}$. Since the width cannot be negative, we take $w = 6$ yards.

Step5: Find the length

Substitute $w = 6$ into the length formula $l=3w - 7$. So $l=3\times6-7=18 - 7 = 11$ yards.

Answer:

Length: 11 yd
Width: 6 yd