by HallsofIvy » Tue Mar 12, 2019 8:09 am
He is named "Mary"? (Probably British!) In any case, let "x" be the width of the garden, in fet so that the length, "5 feet longer than twice its width", is 2x+ 5. "He wants to have a sidewalk along two of the garden’s sides that is 3 feet wide." Which two sides? The two lengths? The two widths? One length and one width?
If the sidewalk is along the two lengths then it is 2 times 2x+ 5 by 3 feet, area 6(2x+ 5)= 12x+ 30= 213 so 12x= 193 and x= 183/12= 15.25 feet. The garden is 15 feet 3 inches wide and 35 feet 6 inches long.
If the sidewalk is along the two widths then it is 2 times x by 3 feet, area 6x= 213 so x= 213/6=35.5 feet. The grarden is 35 feet 6 inches wide and 76 feet long.
If the sidewalk is along one length and one width (most likely) then it is x+ 2x+ 1- 3= 3x- 2 by 3 feet (the "- 3" allows for the overlap at the corner where the two sides meet) so 3(3x- 2)= 9x- 6= 213. 9x= 219 so x= 219/9= 24 feet 4 inches. The garden is 24 feet 4 inches wide and 53 feet 8 inches long.