SOLUTION: The length of a rectangle is 5 inches longer than the width. The perimeter is between 22 and 43 inches. A) Write the perimeter as a compound inequality. B) Write the width as

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 5 inches longer than the width. The perimeter is between 22 and 43 inches. A) Write the perimeter as a compound inequality. B) Write the width as       Log On


   

Question 949916: The length of a rectangle is 5 inches longer than the width. The perimeter is between 22 and 43 inches.
A) Write the perimeter as a compound inequality.
B) Write the width as a compound inequality.
C) Write the length as a compound inequality.
I know that the perimeter is 22 and 34. I don't know how to get the width or the length.
I set up the problem:
x+5+x+5= 22 and 34
2x+10= 22 and 34
2x+10-10= 22-10 and 34-10
2x/2, 12/2, 24/2
6 and 8
Neither were the answer!
Please help!
Thanks.
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
w, width
L, length
L=w+5
p, perimeter bounded by 22 and 43.

2w%2B2L=p
2w%2B2%28w%2B5%29=p
2w%2B2w%2B10=p

The inequality for the perimeter:
22%3C=2w%2B2w%2B10%3C=43
22%3C=4w%2B10%3C=43
22-10%3C=4w%3C=43-10
12%3C=4w%3C=33
Divide all members by 4,
highlight%283%3C=w%3C=8%261%2F2%29

What does that mean for L?
L=3%2B5=8 for one extreme;
L=8%261%2F2%2B5=13%261%2F2 for the other extreme;
highlight%288%3C=L%3C=13%261%2F2%29.