find the dimensions of a rectangle whose length is 5 less than twice its width and whose area is 63 sq. units.

Question

amistre64 · Answer

we know that:

 Area = 63 = lw; where w=w and l=2w-5

63 = (2w-5)w

63 = 2w^2-5w

63 = 2(w^2-5w/2)

63 + 50/16 = 2(w^2-5w/2 + 25/16)

63 + 50/16 = 2(w-5/4)^2

(63 + 50/16)/2 = (w-5/4)^2

sqrt((63 + 50/16)/2) = w-5/4

sqrt((63 + 50/16)/2) + 5/4 = w = 7, so maybe :)

amistre64 · Answer

9*7 = 63  so heres hopin ...

l =2(7) - 5
  = 14 - 5
  =9 ..........yay!!