Question

Area of 320 square feet, the width of the room is 4/5 it's length

Answer 1

Area=legnth times width or
a=lw

if w=4/5l
and a=320 then subsitute 320 for a dn 4/5l for w
320=4/5l times l
multiply both sides by 5 to clear fraction
1600=4l times l
ad like terms
1600=4l^2
subtract 1600 from both sides
0=4l^2-1600
try and factor because if you have xy=0 then x and/or y=0 so
difference of 2 perfect squares where
if you have a^2-b^2 then that equals(a-b)(a+b) so
0=(2l)^2-(40)^2=(2l-40)(2l+40)

so 2l-40=0
add 40 to both sides
2l=40
divide by 2
l=20

if
2l+40=0
subtract 40 from both sides
2l=-40
divide 2
l=-20
since meausrements of real objects cannot be negative, wediscard this number


l=20 is thhe true legnth s
subsitute
w=4/5l
w=4/5(20)
w=80/5
w=16

legnth=20
width=16