The perimeter of a rectangle is 28inches. The length is 1 inch less than twice the width. What is the width?

The answer is 5 inches, as already given, but you might like to know how to do it yourself next time.

Since the question is "what is the width" we will let w be the width.

The length is "one inch less than twice the width" so we will say the length is 2w-1.

The perimeter is equal to 2 widths plus 2 lengths, so we say

p = 2w + 2(2w-1) = 28

I hope you can solve this simple algebraic equation on your own. If not, ask me to solve it and I'll get another 2 points (20 if you pick both answers as best).

The formula for perimeter is 2L+2w For this actual case: 2L+2w=28 that's mentioned that L = 2w-a million by using substitution, 2(2w-a million)+2w=28 4w-2+2w=28 6w-2=28 6w=30 w=5 L=2w-a million Substitution: L=2(5)-a million L=10-a million L=9 The rectangle has a length of 9 inches and a width of 5 inches.

5 in

(1) P=2(W + L)

82) L=2W-1

where P= perimeter, W = width and L is Length.

solving (1) for L:

L=p/2-W

then p/2-W=2W-1

P/2+1 = 2W+W=3W

W=(p/2+1)/3

since P=28

W=(28/2+1)/3=(14+1)/3=15/3=5

and L=2*W-1=10-1=9

then P=2(5+9)=28

W=5

P = 28

l = 2w - 1

P = 2l + 2w

28 = 2(2w - 1) + 2w

28 = 4w - 2 + 2w

28 = 6w - 2

30 = 6w

w = 5 inches

28=2l+2w=>14=l+w

l=2w-1 so

14=2w-1+w=>15=3w=>w=5

i agree 5.

w=5inch

l=9inch