SOLUTION: If the length of a rectangle is 7 meters less than twice the width, we have
L=2W-7
If the area is 60 square meters, we have
L×W=60 substitute L
(2W-7)×W=60
2W^2-7W=60
2W^
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-> SOLUTION: If the length of a rectangle is 7 meters less than twice the width, we have
L=2W-7
If the area is 60 square meters, we have
L×W=60 substitute L
(2W-7)×W=60
2W^2-7W=60
2W^
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Question 1156399: If the length of a rectangle is 7 meters less than twice the width, we have
L=2W-7
If the area is 60 square meters, we have
L×W=60 substitute L
(2W-7)×W=60
2W^2-7W=60
2W^2-7W-60=0
2W^2-15W+8W-60=0
(2W^2+8W)-(15W-60)=0
2W(W+4)-15(W+4)=0
(W+4)(2W-15)=0
2W-15=0 W=15/2 W=7.5
Can someone explain where the 15 and 8 came from? Found 2 solutions by ikleyn, MathTherapy:Answer by ikleyn(52787) (Show Source):
May I give an advise to you ?
The problem is reduced to the quadratic equation
2W^2 - 7W - 60 = 0.
From this point, simply solve it using the quadratic formula.
That is all.
You can put this solution on YOUR website!
If the length of a rectangle is 7 meters less than twice the width, we have
L=2W-7
If the area is 60 square meters, we have
L×W=60 substitute L
(2W-7)×W=60
2W^2-7W=60
2W^2-7W-60=0
2W^2-15W+8W-60=0
(2W^2+8W)-(15W-60)=0
2W(W+4)-15(W+4)=0
(W+4)(2W-15)=0
2W-15=0 W=15/2 W=7.5
Can someone explain where the 15 and 8 came from?
<===== I guess you're referring to the - 15 and + 8 here
The - 15W + 8W REPLACES - 7W since - 15W + 8W = - 7W
The - 15 and + 8 were derived from the NEED to find 2 factors with a product of "ac," or + 2 * - 60 = - 120, and simultaneously SUM to - 7.
These 2 factors are: - 15 and + 8 (notice that the factors have OPPOSITE signs!).
