document.write( "Question 1197396: A rectangular piece of metal is 15 in longer than it is wide. Squares with sides 3 in long are cut from the four corners and the flaps are folded upward to form an open box. If the volume of the box is 1092 in​^3, what were the original dimensions of the piece of​ metal?
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Algebra.Com's Answer #830643 by josgarithmetic(39617)\"\" \"About 
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x, the width of the rectangle piece
\n" ); document.write( "x+15, length of the rectangle piece
\n" ); document.write( "3, HEIGHT of the box formed\r
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\n" ); document.write( "\n" ); document.write( "The base area becomes \"%28x%2B15-2%2A3%29%28x-2%2A3%29=%28x%2B9%29%28x-6%29\".\r
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\n" ); document.write( "\n" ); document.write( "VOLUME is \"highlight_green%283%28x-6%29%28x%2B9%29=1092%29\".\r
\n" ); document.write( "\n" ); document.write( "This may be solvable through checking factoring.\r
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\n" ); document.write( "\n" ); document.write( "\"%28x-6%29%28x%2B9%29=364\"--------try to look for two factors for 364 which differ by 15. Worth a try!\r
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\n" ); document.write( "\n" ); document.write( "\"364=13%2A28\"
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\n" ); document.write( "\"system%28x-6=13%2Cx%2B9=28%29\"---------solve either for x.
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