SOLUTION: A rectangle has a length of 15 inches less than 5 times its width. If the area of the rectangle is 540 square inches, find the length of the rectangle

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Question 1104297: A rectangle has a length of 15 inches less than 5 times its width. If the area of the rectangle is 540 square inches, find the length of the rectangle
Found 3 solutions by Alan3354, josgarithmetic, greenestamps:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A rectangle has a length of 15 inches less than 5 times its width. If the area of the rectangle is 540 square inches, find the length of the rectangle
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L = 5W-15
L*W = 540

Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Dimensions are -15%2B5x and x, for length and width, in that order.

Area was given as 540 square inches.
x%285x-15%29=540
Simplify and solve.

Notice factor of 5 on each side.
x%2A5%2A%28x-3%29=2%2A5%2A54
x%28x-3%29=2%2A54
highlight_green%28x%28x-3%29=108%29
From here, you have two choices. Check for factorizations for 108 which differ by 3; or continue algebra and solve a quadratic equation.

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Both of the other tutors showed the setup for an algebraic solution to the problem.

However, notice that, in the end, you had to do some trial and error to finish solving the problem algebraically.

So if your objective is to find the answer as quickly as possible by any method (as, for example, on a competitive timed test), then trial and error from the beginning might be the best path to the answer.

Simply look for two numbers for the length and width whose product is 540 which satisfy the condition that the length is 15 less than 5 times the width:

10*54 = 540; 5(10)-15 = 35
The value we get for the length using 10 for the width is too small; we need a larger value for the width.

15*36 = 540; 5(15)-15 = 60
Now the value we get for the length using 15 for the width is too big; the width must be between our first guess of 10 and our second guess of 15.

12*45 = 540; 5(12)-15 = 45
AHA! The length of the rectangle is 45.