SOLUTION: Gloria had a rectangular garden plot last year with an area of 60 square feet. This year Gloria's garden plot is one foot wider and three feet shorter than last year's garden, but

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Question 438542: Gloria had a rectangular garden plot last year with an area of 60 square feet. This year Gloria's garden plot is one foot wider and three feet shorter than last year's garden, but has the same area. What are the dimensions of the garden last year? Is the problem (W+1)(L-3)=60 and if so, how do I calculate this?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Gloria had a rectangular garden plot last year with an area of 60 square feet.
This year Gloria's garden plot is one foot wider and three feet shorter than last year's garden, but has the same area.
What are the dimensions of the garden last year?
:
Write an equation for original garden
L*W = 60
L = 60%2FW
Use this for substitution in your new garden equation
(W+1)(L-3) = 60
replace L with 60%2FW
:
(W+1) (60%2FW - 3) = 60
FOIL
60 - 3W + 60%2FW - 3 = 60
:
-3W + 60%2FW - 3 + 60 - 60 = 0
:
-3W + 60%2FW - 3 = 0
:
Multiply eq by W to get rid of the denominator
-3W^2 + 60 - 3W = 0
:
Simplify, divide by -3, change the signs, arrange as a quadratic equation
W^2 + W - 20 = 0
:
Factors to
(W+5)(W-4) = 0
;
Positive solution
W = 4 ft is the original width
then
60%2F4 = 15 ft is the original length
:
:
Check this by finding the area of the new garden
(4+1)*(15-3) =
5 * 12 = 60; confirms our original length of 15, and width of 4