Question 239043: ok i really need help on this problem. i am in honors algebra 2 and i dont know what to do. Please help me solve the word problem: A rectangle is 3 times as long as it is wide. If the length and the width are both decreased by 4 cm, the resulting rectangle has area 28cm squared. Find the dimensions of the original rectangle.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! A rectangle is 3 times as long as it is wide. If the length and the width are both decreased by 4 cm, the resulting rectangle has area 28cm squared. Find the dimensions of the original rectangle.
L = 3W
(L-4) * (W-4) = 28
Replace L with 3W to get:
(3W-4) * (W-4) = 28
remove parentheses by multiplying out the factors to get:
3W^2 - 16W + 16 = 28
subtract 28 from both sides to get:
3W^2 - 16W - 12 = 0
this factors out to:
(3W+2) * (W-6) = 0
This makes 3W = -2 or W = 6
W can't be negative so W = 6 is the only valid answer.
if W = 6, then L = 3W makes L = 18
You have L = 18 and W = 6
substitute in the original equation to get:
(L-4) * (W-4) = 28 becomes:
(18-4) * (6-4) = 28 which becomes:
14 * 2 = 28 which becomes:
28 = 28 which is true so the values for L and W must be good.
Your answers are:
L = 18 and W = 6
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