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Question 104718: Hi, I have been struggling all morning with a problem. Here it is:
The length of a rectangle is 1 less than twice its width. If the width is 3 more than one-seventh of the perimeter, find the dimensions of the rectangle.
While I can come up with a value for P which is a term, not a value, I cannot come up with any actual dimensions, and my answer will not proof out. I have approached the problem different ways, such as width=x and length=2x+1 (using the 2L = 2S = P standard equation for finding perimeter), and w=1/7 P - 3, etc. for length, but I cannot come up with an actual answer. Thanks in advance.
Maddie
Found 2 solutions by doukungfoo, scott8148:
Answer by doukungfoo(195)   (Show Source):
You can put this solution on YOUR website! Perimeter of a rectangle is equal to the sum of its four sides.
A rectangle has two sides that represent its length and
two sides that represent its width. So
perimeter = 2 times length plus 2 times width
P = 2L + 2W
Now for this rectangle we are told that the
length = 2 times width minus 1
L = 2W - 1
We are also told that the
width = 1/7 perimeter plus 3
W = 1/7(P) + 3
W = P/7 + 3
Ok lets go back to the original equation for perimeter
P = 2L + 2W
since we have shown that L = 2W - 1 we can substitute 2W - 1 for L in
P = 2L + 2W
P = 2(2W-1) + 2W
Now if we use this equation with W = P/7 +3 we have a system of equations in two varibles that we can solve for.
So here are the two equations we will be working with:
P = 2(2W-1) + 2W
AND
W = P/7 + 3
Lets solve the system of equations using the substitution method.
Since W = P/7 + 3 we can substitute P/7 + 3 for the W in
P = 2(2W-1) + 2W
P = 2(2(P/7 + 3)-1) + 2(P/7 + 3)
Now we have and equation with just one variable to solve for
P = 2(2(P/7 + 3)-1) + 2(P/7 + 3)
P = 2(2P/7 + 6 - 1) + 2P/7 + 6
P = 4P/7 + 12 - 2 + 2P/7 + 6
P = 6P/7 + 16
P = 6P/7 + 112/7
7P = 6P + 112
P = 112
The perimeter is 112
use this to find the width
W = P/7 + 3
W = 112/7 + 3
W = 16 + 3
W = 19
The width of the rectangle is 19
use this to find the length
L = 2W - 1
L = 2(19) - 1
L = 38 - 1
L = 37
The length of the rectangle is 37
check answers with formula for perimeter
P = 2L + 2W
112 = 2(37) + 2(19)
112 = 74 + 38
112 = 112
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! a couple of your equations have sign problems
let x=width
"The length of a rectangle is 1 less than twice its width" ... length=2x-1
"the width is 3 more than one-seventh of the perimeter" ... (p/7)+3=x ... p=7(x-3)
2(x)+2(2x-1)=7(x-3)
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