length of rectangle is 3 more than three times its width perimiter is 62 find length and width
2006-09-10 04:34:58 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
let x=the rectangle's width
2x+2(x+3)=62
2x+2x+6=62
4x=56
x=14
width=14, length=17
2006-09-10 04:38:30 · answer #1 · answered by cardsfan 2 · 0⤊ 2⤋
P=2l +2w
from the problem l=3w+3
substitute back into perimeter formula
62=2(3w+3) + 2w
62=6w + 6 + 2w
56=8w
7=w
to find l sustitute 7 for w in the length formula
l=3w + 3
L= 3(7) + 3
= 24
2006-09-10 11:41:24 · answer #2 · answered by absynthian 6 · 0⤊ 0⤋
7X24.
You have six extra due to the +3, so get rid of that and you have 56. Since each pair of one side and one end is half of 56, each end is an eighth of 56, which is 7. (3*7)+3=24, the length of the side. Just to check it, (24+7)*2=31*2=62, so your answer is 7 by 24.
2006-09-10 11:39:06 · answer #3 · answered by Anonymous · 1⤊ 0⤋
Let l be the length and w the width.
Then 3w + 3 = l
and 2w + 2l = 62.
Rewriting these,
3w - l = -3
w + l = 31
So
4w = 28
w = 7
l = 24.
2006-09-10 11:52:02 · answer #4 · answered by steiner1745 7 · 0⤊ 0⤋