# SOLUTION: Hello, I need help with this problem: The length of a picture is one inch less than twice its width. The frame around the picture has a uniform width of two inches and an

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 Click here to see ALL problems on Miscellaneous Word Problems Question 90934: Hello, I need help with this problem: The length of a picture is one inch less than twice its width. The frame around the picture has a uniform width of two inches and an area of 96 square inches. What are the dimensions of the picture?Answer by ankor@dixie-net.com(15652)   (Show Source): You can put this solution on YOUR website!The length of a picture is one inch less than twice its width. The frame around the picture has a uniform width of two inches and an area of 96 square inches. What are the dimensions of the picture? : Let x = width of the picture then (2x-1) = length of the picture : It would help to draw a rough diagram of this. Label the picture dimensions as x and (2x-1). Label the uniform width around the picture as 2 inches. It will be apparent that the overall dimensions of the frame will be 2x+3, (from 2x-1+4) and x+4. : The overall dimensions would be (2x+3) by (x+4), it's area is given as: 96 sq/in Length time width = area. Therefore: (2x + 3)*(x+4) = 96 FOIL 2x^2 + 11x + 12 = 96 : 2x^2 + 11x + 12 - 96 = 0 : 2x^2 + 11x - 84 = 0; a quadratic equation : Use the quadratic formula to find x: a=2; b=11; c=-84 : : : ;minus a minus is a plus : : ; we only want the positive solution here : : x = 4.29 inches : Find the dimension of the picture: Length: 2(4.29) - 1 = 7.58 inches Width = 4.29 inches : : Check our solution by finding the frame area: (7.58+4) * (4.29+4) = 11.58 * 8.29 = 95.998 ~ 96 sq/in, the given area : : Did this make sense to you? Any questions?