The length of the famed fred's fence is 5 more than twice it's width if the perimeter of the fence is 510 feet find the dimensions of his fenced in area

Step-by-step explanation:

Let w represent the width of the area. Then 2w+5 represents the length. The perimeter is twice the sum of length and width:

  P = 2(L+W)

  510 = 2(2w+5 +w) . . substitute given values

  255 = 3w +5 . . . . . . divide by 2

  250 = 3w . . . . . . . . . subtract 5

  83 1/3 = w . . . . . . . . .divide by 3; the width of the area

  2w+5 = 171 2/3 . . . .  the length of the area

The fenced area is 83 1/3 ft wide and 171 2/3 ft long.

35 is an outlier

step-by-step explanation:

or

step-by-step explanation:

we need to solve for 'x' the following equation:  

we have that:

⇒

⇒ ⇒

factorizing, we get that:

  ⇒

then, the solutions are: or

nothing extraneous has been found!