Perimeter of swimming pool is 58m. Length is 11m more than the width. Find the dimensions. HELP!!?

Question

Here's the question in it's entirety.

A construction company builds a swimming pool with a perimeter of 58m. The length is 11m more than the width. Find the dimensions of the swimming pool.

Answer 1

width = x
length = x + 11

2l + 2w = p
2(x + 11) + 2x = 58
2x + 22 + 2x = 58
4x + 22 = 58
4x = 36
x = 9

width = 9
length = 20

Answer 2

this is a frequent question on Yahoo answer's. it's just that the details are different: the numbers may change, and it may be a fence around a plot of land, or something else. but the way to solve it is always the same.

You are given some perimeter P, and told that the length L is some value x greater than the width W. So P and x are known and we must find L and W.

Since it is a rectangle, the perimeter is:

P = 2(L+W)
L=W+x

here we have two equations in two unknowns, which can be solved several ways. One way is simple substitution: since L=W+x, replace L with W+x in the first equation.

You get
P = 2(W+x+W) = 2(2W+x)
solving for W, we get
W = (P/4) - (x/2)
and using this value we can then find L
L = W+x = (P/4) - (x/2) + x = (P/4) + (x/2)

thus we have the general solution for such problems:
W = (P/4) - (x/2)
L = (P/4) + (x/2)

for this problem P=58 and x=11
P/4= 58/4 =14.5
and x/2 = 11/2 = 5.5

so we have 14.5 +/- 5.5 = 20 and 9 for L and W, respectively.

Answer 3

Dead easy.

perimeter is 58.
Length is 11 m more than width so
2 x 11 = 22
58 - 22 = 36
So width is 36/4 = 9
Length is 9 + 11 = 20

2 x 20 = 40
2 x 9 = 18
40 + 18 = 58

Answer 4

58 = 2L + 2W
L=W+11

58=2(W+11) + 2W
58= 2W+22+2W
58=4W +22
36=4W
9=W

L=9+11
L=20

20+9+20+9 =58