Question 161047This question is from textbook Amsco's Integrated Algebra 1
: Hi I really need help on my math homework. Here is the question: The length of a rectangle exceeds its width by 4 feet. If the width is doubled and the length is decreased by 2 feet, a new rectangle is formed whose perimeter is 8 feet more than the perimeter of the original rectangle. Find the dimensions of the original rectangle.
My teacher says that we have to do "let" statements first. Her are mine:
Let x= the width of the rectangle
Let x + 4= the length of the rectangle
I am not sure if these let statements are right, and if they are i am not sure how to set up the equation. Thanks for help
This question is from textbook Amsco's Integrated Algebra 1
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website!
YOU ARE ABSOLUTELY RIGHT IN YOUR APPROACH SO FAR:
Now given what you have done, we know that the perimeter of a rectangle is 2w+2l
So, the perimeter of the original rectangle is:
2x+2(x+4)
Now if we double the width, we get: 2x
And if we decrease the length by 2, we get: x+4-2=x+2
Now the new perimeter would be 2*2x+2(x+2)=4x+2(x+2) and we are told that this new perimeter is 8 feet more than the perimeter of the original rectangle, so our equation to solve is:
Original perimeter +8 =new perimeter, or
2x+2(x+4)+8=4x+2(x+2) get rid of parens (distributive law)
2x+2x+8+8=4x+2x+4 collect like terms
4x+16=6x+4 subtract 4x and also 4 from each side
4x-4x+16-4=6x-4x+4-4 collect like terms again
12=2x divide each side by 2
x=6 ft--------------------------width of the original rectangle
x+4=6+4=10 ft----------------------length of the original rectangle
CK
perimeter of original rectangle=2*6+2*10=32 ft
perimeter of new rectangle=2*12+2*8=40 ft
40-32=8
8=8
Hope this helps----ptaylor
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