SOLUTION: I really need help with this and need this word problem translated into an algebraic equation with only one variable. Then it needs to be simplified.
Question:
The area of a rect
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Question 505410: I really need help with this and need this word problem translated into an algebraic equation with only one variable. Then it needs to be simplified.
Question:
The area of a rectangle is 72 square inches. If the length is 40 inches greater than four times the width, what are the dimensions of the rectangle?
Found 2 solutions by ankor@dixie-net.com, Taxie:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The area of a rectangle is 72 square inches.
L * W = 72 sq/in
:
If the length is 40 inches greater than four times the width,
L = 4W+40
:
In the 1st equation, replace L with (4W+40)
(4W+40) * W = 72
4W^2 + 40W = 72
Simplify, divide by 4
W^2 + 10W = 18
A quadratic equation
W^2 + 10W - 18 = 0
Won't factor, have to solve it with the quadratic formula
a=1; b=10; c= -18
The positive solution = 1.55743852" is the width of the rectangle
then
4(1.55743852) + 40 = 46.22975408" is the length
:
Confirm this by finding the area using these two dimensions
Answer by Taxie(4) (Show Source): You can put this solution on YOUR website!
call the width w. The length is then 4w+40
The area of the rectangle is then. 72 = 4w + 40
Simplify dividing the equation by 4 to get. 18 = w + 10
The solution is w = 8
(sorry I misread 4 for 40 in th previous post)
Ciao
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