SOLUTION: what is the dimension of a rectangle whose length is 5 less than twice its width and whose area is 63 square units?

Question 581757: what is the dimension of a rectangle whose length is 5 less than twice its width and whose area is 63 square units?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
what is the dimension of a rectangle whose length is 5 less than twice its width and whose area is 63 square units?
.
Let w = width
then from: "length is 5 less than twice its width" we get
2w-5 = length
.
since:
area = width * length
63 = w(2w-5)
63 = 2w^2-5w
0 = 2w^2-5w-63
rewrite the middle term as:
0 = 2w^2-14w+9w-63
group:
0 = (2w^2-14w) + (9w-63)
0 = 2w(w-7) + 9(w-7)
0 = (w-7)(2w+9)
w = {-9/2, 7}
toss out the negative term (extraneous) leaving:
w = 7 units (width)
.
Length:
2w-5 = 2(7)-5 = 14-5 = 9 units
.
Dimension: 7 units by 9 units

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