Simple Math Question.

[ZenMasterFlex]

The Perimeter of a rectangular football field 230 ft.
If 2 times the width is 5 feet less than the length what is the width?

What formula do you use to solve this?

Answer please.

[Psycho Boy Jack]

P=230=2L+2W
2W=L-5
P=2L+L-5
P=3L-5

3L-5=P then L=(P+5)/3 and W=(((P+5)/3)-5)/2
W=73.3333333/2=36.6

[Sephus]

lol at this being posted here.

[MisterW]

width, W = 36'-8"
length, L = 78'-4"

2 equations, 2 variables, solve

2*W + 2*L = 230'-0"
2*W = L - 5

Pick an equation, solve for L in terms of W or W in terms of L (for example, rearrange equation 2 such that L = 2*W +5)

Substitute the variable you solved for into the other equation (for example, replace L in equation 1 with 2*W + 5)

Solve for W, then use this value in one of the equations to solve for L

[Sephus]

hehe nice edit.

[Psycho Boy Jack]

lol.
i forgot a /2 only in the numerical expression..... cmon.
wanted to be the first so did it in ten seconds [img]/images/graemlins/smile.gif[/img]

[ZenMasterFlex]

o.k.... here is the exact problem as stated:

The length of a rectangular playing field is 5 feet less than twice its width. If the perimeter of the playing field is 230 feet, find the length and width of the playing field.

Just to make sure it is the same.

[Psycho Boy Jack]

you have the results just above... well it is letters
L lenghth
W width
perimeter = 2W+2L
and 2W=L-5
so (1) becomes P=L-5+2L=3L-5
so L=P+5)/3
and as you know from 2 that 2W=L-5
then W=(L-5)/2=(((P+5)/3)/2)
numerical: 36.6666666 and 78.33333333

[Sephus]

[ QUOTE ]
o.k.... here is the exact problem as stated:

The length of a rectangular playing field is 5 feet less than twice its width. If the perimeter of the playing field is 230 feet, find the length and width of the playing field.

Just to make sure it is the same.

[/ QUOTE ]

lolo

[MisterW]

Not the same. In this case, L = 2*W-5.

W = 40'-0"
L = 75'-0"