Question

the length of a rectangle is 11 yd less than three times the width, and the area of the rectangle is 42 yd². find the dimensions of the rectangle.

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ yards. Then the length $l = 3w - 11$ yards.

Step2: Set up area - formula equation

The area of a rectangle is $A=l\times w$. Given $A = 42$, we substitute $l$ and $A$ into the formula: $(3w - 11)w=42$.

Step3: Expand the equation

Expand $(3w - 11)w$ to get $3w^{2}-11w = 42$. Rearrange it to the standard quadratic - form $3w^{2}-11w - 42 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 3$, $b=-11$, $c = - 42$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ or factor. Factoring $3w^{2}-11w - 42=(3w + 7)(w - 6)=0$.
Setting each factor equal to zero gives $3w+7 = 0$ or $w - 6=0$.
From $3w+7 = 0$, we get $w=-\frac{7}{3}$, but since the width cannot be negative, we discard this solution.
From $w - 6=0$, we get $w = 6$ yards.

Step5: Find the length

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

Answer:

Length: 7 yd
Width: 6 yd