The area of a rectangle is 65yd^2 , and the length of the rectangle is 3yd less than double the width, how do you find the dimensions?

1 Answer

Answer:

build the equations and solve...

Explanation:

let the area be #A = l * w# where length is #l# and width is #w#

so 1.st equqtion will be

#l*w=65#

and length is 3 yd less than double the width says:

#l = 2w-3# (2nd eq.)

replacing #l# with #2w-3# in first eq. will yield

#(2w-3) * w = 65#
#2w^2-3w = 65#
#2w^2-3w-65 = 0#

now we have a 2nd order equation just find the roots and take the positive one as width can not be negative...

#w=(3+-sqrt(9+4*2*65))/(2*2) = (3+-sqrt(529))/(4) = (3+-23)/4 #
#w=-5 , 13/2# so taking #w=13/2 = 6.5 yd#

replacing #w# with #6,5# in second eq. we get

#l=2w-3 = 2*6.5-3 = 13-3 = 10 yd#

#A = l*w = 10 * 6.5 = 65yd# will confirm us...