The perimeter of a college athletic field is 100 meters and the length is 18 m more than the width. How do you find the length and width. How would I find the answer?

I would form an algebraic expression from the information given:

Let length of field be represented by #x# and width by #y#.

We are told that perimeter is 100m, so #2x+2y=100#
which simplifies to:
1. #x+y=50#
by dividing both sides by 2

The other info tells us that length,#x# is 18m more than width,#y#:
2. #=>x=y+18#

Then solve the equations simultaneously (I have used substitution):
#(y+18)+y=50#
#=> 2y= 50-18#
#=>y=32/2#
#=>y=16#

Then substitute back into either equation to obtain #x#:
2. #x=16+18 = 34#

Double check by making sure #2x+2y=100#

Hope this helps!

#"Given that:"#

#"Perimeter " ##=100##" meters"#

#"Let us assume the width as "##w##" meters."#

#"So the length "##=w+18##" meters"#

#"According to question:"#

#"Perimeter"##=2(l+w)#

#100=2(w+w+18)#

#100/2=w+w+18#

#50=2w+18#

#50-18=2w#

#32=2w#

#w=32/2#

#color(red)(w"=16 meters"#

#"So, we have found that the width is 16 meters."#

#"Length"=w"+18"#

#=16+18#

#color(red)("=34 meters"#

#color(magenta)("The width of the field is 16 meters and the length is 34 meters"#

#color(darkred)("As a check:-"#

Perimeter=#2(l+w)#

#100=2(34+16)#

#RHS= 2(34+16)#

#=2xx50#

#=100#

#LHS=100#

#"Since LHS = RHS", #

#"Hence it is proved that the length = 34 meters and the width = 16 meters."#

#"Hope this helps! :)"#