SOLUTION: The length of a picture without its border is 7 inches less than twice its width. If the boarder is 1 inch wide and the area is 62 square inches, what are the dimensions of the pi
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Question 320662: The length of a picture without its border is 7 inches less than twice its width. If the boarder is 1 inch wide and the area is 62 square inches, what are the dimensions of the picture alone?
2X-7+1=62
6-2X=62
+6= +6
=68/2
34= 8.5 BY 4
You can put this solution on YOUR website! The length of a picture without its border is 7 inches less than twice its width. If the boarder is 1 inch wide and the area is 62 square inches, what are the dimensions of the picture alone?
:
Frankly, I haven't a clue on what you're doing here, so starting from scratch:
:
x = the width of the picture
and
(2x-7) = the length of of the picture
then
Area of the picture: x(2x-7) = 2x^2 - 7x
:
The border and the picture (2" will be added to the picture dimensions)
(x + 2) = overall width
and
2x - 7 + 2 = (2x-5) = overall length
then
overall area = (x+2)*(2x-5) = 2x^2 - 5x + 4x - 10 = (2x^2 - x - 10)
:
overall area - picture area = border area
(2x^2 - x - 10) - (2x^2 - 7x) = 62
remove the brackets
2x^2 - x - 10 - 2x^2 + 7x = 62
Combine like terms
2x^2 - 2x^2 - x + 7x = 62 + 10
:
6x = 72
x =
x = 12" is the width of the picture
then
2(12) - 7 = 17" is the length of the picture
:
:
Check solution by finding the area of the border
(2" added to the overall dimensions)
19*14 - 17*12 =
266 - 204 = 62 sq/in
:
:
How about this? Could you follow this OK?
