Question 935055: The length of a rectangle is 7yds
less than three times the width, and the area of the rectangle is 66yds^2
. Find the dimensions of the rectangle.
Found 3 solutions by TimothyLamb, josgarithmetic, MathTherapy:
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! L = 3w - 7
Lw = 66
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Lw = 66
(3w - 7)w = 66
3ww - 7w - 66 = 0
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the above quadratic equation is in standard form, with a=3, b=-7 and c=-66
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
3 -7 -66
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two real roots at:
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w = 6
w = -3.6666666
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the negative root doesn't fit the problem statement, so use the positive root:
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w = 6
L = 3w - 7
L = 3*6 - 7
L = 11
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answer:
w = 6 yds
L = 11 yds
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Answer by josgarithmetic(39618)  (Show Source):
You can put this solution on YOUR website! L for length, w for width.
A for area.
L and w are unknown.
The length of a rectangle is k yards
less than m times the width w, and the area of the rectangle is A yds^2.
Find the dimensions, w and L.
 , 
Beginners would rely on the quadratic expression being factorable when m, k, and A are substituted,
and then solution for w would be found that way (factoring). Intermediate level and higher students
could solve for w either by quadratic formula solution, or competing the square.
Directly using general solution,
One of those will make sense but not the other. Substitute the values for k, m, and A, evaluate w,
and then use the value to evaluate L.
Answer by MathTherapy(10552)  (Show Source):
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