The given operation involves just swapping the two rows, so carrying out
on the matrix gives
![\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%265%265%5C%5C1%26-%5Cdfrac23%265%5Cend%7Barray%7D%5Cright%5D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26-%5Cdfrac23%265%5C%5C5%265%265%5Cend%7Barray%7D%5Cright%5D)
Let the width be x.
Length is 8 feet more than width. Length = x + 8
Area = x(x + 8)
width increased by 4, that is, (x + 4)
Length decreased by 5, (x + 8 - 5) = (x + 3)
Area = (x + 4)(x +3)
Area remains the same
x(x + 8) = (x+4)(x +3)
x² + 8x = x(x +3) + 4(x +3)
x² + 8x = x² +3x + 4x +12
x² + 8x = x² +7x +12 Eliminate x² from both sides
8x = 7x + 12
8x - 7x = 12
x = 12
Dimensions of original rectangle : x, x + 8
12, 12 +8 = 12, 20
Original rectangle is 20 feet by 12 feet
It is likely D I am not positive
I believe the answer would be 1620
The formula for the area of a triangle is (base•height)/2
So 60(54) = 3240/2 = 1620
Please correct me if I am wrong have a good day!
