SOLUTION: Could someone be so kind in helping me with this problem?
I need to solve the following problem and show the equation used for the solution:
The length of a rectangle is
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Question 58421This question is from textbook Beginning Algebra
: Could someone be so kind in helping me with this problem?
I need to solve the following problem and show the equation used for the solution:
The length of a rectangle is 2 in. more than twice its width. If the perimeter of the rectangle is 34 in., find the dimensions of the rectangle.
Thank you,
Lisa
This question is from textbook Beginning Algebra
Answer by mszlmb(115) (Show Source): You can put this solution on YOUR website!
The Permimeter=2L+2W. We know our P is 34 ["Perimeter of the rectangle is 34 in,"].
The length is two more than twice the width.
Twice the width is 2w.
Two more than that is 2w+2.
Therefore, the length (l)=2w+2
Back to the first equation, 34=2L+2W. We can replace the L in 2L with (2w+2) since L=(2w+2) Second equation. Let's:
34=2w+2(2w+2) Distribute
34=2w+4w+4 Minus 4 both sides, combine like terms
30=6w Divide both sides by 6
5=w. There, the width is 5.
Plug that in to the second equation (l=2w+2) to obtain l:
l=2(5)+2
l=10+2
l=12 The length is 12!
Let's double check:
2(12)+2(5)=34
24+10=34
34=34 We're good so far!
2(5)+2=12
10+2=12
12=12
Yea we got it right! For questions or comments email or IM me.
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