Question

hey guys I have intense need of help with two pooptastic word problems. Anyone feel up to it??????hu huhuhu? I know you want to solve it....
The dimensions of a rectangle are such that its L is 5 inches more that its W. If the L were doubled and the W were decreased by 2, the area would be increased by 162 inches. What are the L and W of the rectangle?

Answer 1

this is actaully a repeat from the other day. is this an on line class?

Answer 2

Is it?

Answer 3

where is the first repeat lol

Answer 4

Wow

Answer 5

WOW?

Answer 6

can someone clue me in I feel like im the odd man ?woman out:(

Answer 7

L = 5 +W
(2L)(W-2) = 162

2L =162/w-2
L = 81/w-2

5+ W = 81/w-2

5w - 10 + w^2 - 2w = 81

w^2 + 3w -10 = 81

(w+5)(w-2) = 81

w = 76
w = 84

L = 81 when width is 76
L = 89 when width is 81


am i right?

Answer 8

in any case since L is 5 inches more than W, we can write
\[L=W+5\] and then double L to get
\[2(W+5)=2W+10\] decrease W by 2 to get
\[W-2\] and then write
\[W(W+5) \] as the first area,
\[(W-2)(2W+10)\] as the second area and we have
\[W(W+5)+162=(W-2)(2W+10)\] and solve for W

Answer 9

You guys make me fel so inadequate....thankzxxx

Answer 10

ii get
\[W=13\] as an answer hmm

Answer 11

maybe i did something wrong, let me see

Answer 12

if i write it with paper and pencil maybe i won't mess up