Question 406058: Michell went into a frame shop. He wanted a frame that was 3 inches longer than it was wide. The frame he chose to extend 1.5 inches beyond the picture on each side. Find the outside dimensions of the frame in the area of the unframed picture in 70in^2.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Michell went into a frame shop. He wanted a frame that was 3 inches longer than it was wide.
The frame he chose to extend 1.5 inches beyond the picture on each side.
:
I am going to change this sentence to:
"Find the outside dimensions of the frame if the area of the unframed picture is 70 in^2."
If this is inaccurate, then I don't understand the problem, ignore what follows.
:
Let x = the width of the frame
then
(x+3) = the length of the frame
:
If the frame is 1.5 inches larger than the picture all the way around, then
x - 3 = the width of the picture
and
x + 3 - 3 = x is the length of the picture
:
The picture area
x(x-3) = 70
x^2 - 3x = 70
x^2 - 3x - 70 = 0
Factors to:
(x-10)(x+7) = 0
The positive solution is what we want here
x = 10 inches is the width of the frame
ten
10 + 3 = 13 inches is the length of the frame
:
The frame is 13" by 10"
:
:
:
Check this: the picture dimensions will be 3" less than the frame, therefore:
10 * 7 = 70; the given area of the picture
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