(L-4)L=area=(L+8)(w-4)
(L-4)L=(L+8)(L-4-4)
L^2-4L=L^2-8L +8L-64 check that.
-4L=-64
solve for l, then W
the length of a rectangle is 4cm more than the width of the rectangle. If you increase the length by 8cm and decrease the width by 4cm, the area will remained unchanged. Find the original dimensions
2 answers
old width ----x
old length --- x+4
new width --- x-4
new length --- x+4 +8 =x + 12
x(x+4) = (x-4)(x+12)
x^2 + 4x = x^2 + 8x - 48
-4x = -48
x = 12
old rectangle is 12 by 16
for an area of 192
new rectangle is 8 by 24
for an area of 192
My answer is correct
old length --- x+4
new width --- x-4
new length --- x+4 +8 =x + 12
x(x+4) = (x-4)(x+12)
x^2 + 4x = x^2 + 8x - 48
-4x = -48
x = 12
old rectangle is 12 by 16
for an area of 192
new rectangle is 8 by 24
for an area of 192
My answer is correct