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.
:
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