SOLUTION: The area of a rectangle is 15m^2 + 11m - 12. Find the length if the width is 5m-3

Question 169630: The area of a rectangle is 15m^2 + 11m - 12. Find the length if the width is 5m-3
Found 2 solutions by checkley77, gonzo:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
15m2+11m-12
(5m-3)(3m+4)
3m+4 is the length.
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
area of a rectangle equals length * width.
let A = area
let L = length
let W = width
A = L*W
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you are given the area and the width.
you want to find the length
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A = L*W
divide both sides of equation by W:
A/W = L
which is the same as:
L = A/W
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A = 15m^2 + 11m - 12
W = 5m-3
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substitute for A and W:
L = (15m^2 + 11m - 12) / (5m-3)
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this is a straight long division problem:
first you figure how many times 5m will divide into 15m^2.
answer is:
3m ********* this is the first part of your answer.
then you multiply (5m-3) * 3m
answer is:
(15m^2 - 9m)
then you subtract (15m^2 - 9m) from (15m^2 + 11m - 12)
answer is:
(15m^2 + 11m - 12) - (15m^2 - 9m) =
15m^2 + 11m - 12 - 15m^2 + 9m =
(20m - 12)
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next you figure out how many times 5m goes into 20m.
answer is:
4 ********* this is the second part of your answer
then you multiply (5m-3) * 4
answer is:
(20m - 12)
then you subtract (20m-12) from (20m - 12)
answer is:
20m - 12 - (20m - 12) =
20m - 12 - 20m + 12 =
0
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your answer for (15m^2 + 11m - 12) / (5m-3) is (3m + 4)
(3m+4) is the length.
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to prove your answer is correct, go back to the original equation of:
A = L*W
substitute for A, L, and W:
15m^2 + 11m - 12 = (3m+4)*(5m-3)
(3m+4)*(5m-3) = 15m^2 - 9m + 20m - 12 =
15m^2 + 11m - 12.
this is the same as the area you started with so the equation is true proving that the value of (3m+4) for the length is good.
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your answer is:
length = (3m+4)
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