Arectangle has a perimeter of 204 feet. it's length is six feet longer than twice it's width. if l stands for the length of the rectangle and w stands for its width, write a system of equations that models the information given in this problem and solve it to find the length and width of this rectangle

L = 70

W = 32

Step-by-step explanation:

P = 2L + 2W

P 204 feet

W = x

L  = 2x + 6

P = 2L + 2W

P = 2*(2x + 6) + 2*x

P = 4x + 12 + 2x

P = 6x + 12

But we know that P = 204  So 6x + 12 = 204

6x + 12 = 204                    Subtract 12 from both sides

6x + 12 - 12 = 204 - 12       Collect like terms on the right

6x = 192                             Divide by 6

x = 192/6

x = 32

Answer

W = 32

L = 2*W + 6

L = 2*32 + 6

L = 64 + 6

L = 70

the answer is (2x+8)²= 16

step-by-step explanation: