Question

1. a rectangle has a length that is three feet more than twice its width. (a) if the width of the rectangle is represented by the variable w, give an expression for the length of the rectangle in terms of w. (b) what would the perimeter of this rectangle be if the width was 10 feet? (c) the perimeter of the rectangle is 78 feet. set up an equation using w that models this situation. solve it and state the values for both the width and the length of the rectangle.

Explanation:

Step1: Find length expression

Given length is three feet more than twice width. If width is $w$, then length $l = 2w+3$.

Step2: Find perimeter formula

Perimeter of rectangle $P = 2(l + w)$. Substitute $l = 2w + 3$ into it, we get $P=2((2w + 3)+w)=2(3w + 3)=6w+6$.

Step3: Solve for width when perimeter is 78

Set $P = 78$ in $P=6w + 6$. So, $78=6w+6$.
Subtract 6 from both sides: $78 - 6=6w$, which gives $72 = 6w$.
Divide both sides by 6: $w=\frac{72}{6}=12$ feet.

Step4: Solve for length

Substitute $w = 12$ into length formula $l = 2w+3$. Then $l=2\times12 + 3=24 + 3=27$ feet.

Answer:

(a) $l = 2w+3$
(b) 66 feet
(c) Width $w = 12$ feet, Length $l = 27$ feet