The length of a rectangle is 3 feet more than twice the width. if the width is increased by four feet the perimeter of the new rectangle is 68 feet. find the area of the original rectangle

The area of the original rectangle  = 189 sq ft.

Step-by-step explanation:

Let, the width of the rectangle = m feet

So, the length of the rectangle =  (2 m + 3)  ft

Now, new width w'  = (m + 4)

Perimeter of new rectangle (P')   = 68 ft

Perimeter of a rectangle = 2 (LENGTH + WIDTH)

⇒68 ft   = 2( L + W')  

or, 68 ft   = 2 [( 2m + 3) + (m + 4)]

or, 2 (3m + 7)  = 68

or, 3m + 7 = 68/2  = 34

⇒ 3m = 34 - 7 = 27

⇒ m = 27/3 = 9

or, m = 9 ft

Hence, the original width  of the rectangle =  9 ft

Original Length of the rectangle = 2m  + 3  = 2(9) + 3 = 21 ft

Now, Area of the Rectangle = LENGTH X WIDTH

=  21 ft  x  9 ft  = 189 sq ft

Hence,the area of the original rectangle  = 189 sq ft.

hi,

6x times 4 is 24x and 5 times 4 is 20, so the answer is 24x+20.

hope this and for a little humor i am groot.