SOLUTION: The length of a rectangle is 3 less than 5 times of its width. write a simplified algebraic expression for the perimeter of a rectangle. if the rectangle width is tripled and its

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Question 957612: The length of a rectangle is 3 less than 5 times of its width.
write a simplified algebraic expression for the perimeter of a rectangle.
if the rectangle width is tripled and its length is doubled the perimeter of new rectangle is 92 cm greater than original perimeter.
Find the area of the original rectangle.
PLZZZ explain how u solved this?
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
W=width; L=length=5W-3; P=perimeter
Original rectangle:
P=2(L+W) Substitute for L
P=2((5W-3)+W)=6W-3=12W-6
Original perimeter is 12W-6
New rectangle:
W%5Bn%5D=3W; L%5Bn%5D=2L=10W-6
=2%2813W-6%29=26W-12
New perimeter is 26W-12
New perimeter-original perimeter=92 cm
P%5Bn%5D-P=92cm
%2826W-12%29-%2812W-6%29=92cm
14W-6=92cm
14W=98cm
W=7cm The original width was 7cm.
L=5W-3=5(7cm)-3=35cm-3cm=32cm The original length was 32cm.
Area=L*W=32cm*7cm=224 sq cm
ANSWER: The area of the original rectangle is 224 square centimeters.
CHECK
New Perimeter-Original Perimeter=92cm
2(2L+3W)-2(L+W)=92cm
2(64cm+21cm)-2(32cm+7cm)=92cm
2(85cm)+2(39cm)=92cm
170cm-78cm=92cm
92cm=92cm