Afarmer is enclosing a rectangular area for a pigpen. he wants the length of the pen to be 15 ft longer than the width. the farmer can only afford no more than 150 ft of fencing. what is the pen's greatest possible length? explain your answer and you..

45 ft

Explanation:

Let 

x = width of the pigpen
x + 15 = length of the pigpen (since the length is 15 ft longer than the width)

Since the pigpen is enclosed in a rectangle, the total length of the fence is equal to the perimeter of the pigpen. Moreover, since the farmer can't afford more than 150 ft of fencing, the total length of the fence is less than or equal to 150 ft. So, the perimeter of the pigpen is less than or equal to 150. 

Since the pigpen is rectangular

Perimeter = 2[(length) + (width)]
                = 2[(x + 15) + x]
                = 2(2x + 15)
Perimeter = 4x +30     (1)

Since the perimeter is less than or equal to 150, using equation (1),

4x + 30 ≤ 150
4x + 30 - 30 ≤ 150 - 30
4x ≤ 120
x ≤ 30    (divide both sides by 4)

Hence,

length = x + 15 ≤ 30 + 15 = 45

So, the length cannot be more than 45 ft. Therefore, the greatest possible length is 45 ft.  

1. isolate the variable, in this case, d. subtract 35.5 on both sides of the equation.2. 52 subtracted by 35.5 is 16.53. therefore your answer would be d= 16.5this is the basics of pre algebra, isolating variable through adding, subtracting, multiplying, and dividing, if you have any other questions to ask just comment!