document.write( "Question 1104022: An oil painting is 10 inches longer than it is wide and is bordered on all sides by a 3 inch wide frame. If the area of the frame alone is 402 inches squared, what are the dimensions of the painting? \n" ); document.write( "

Algebra.Com's Answer #718745 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "The problem says\r\n" );
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document.write( "L - W = 10,                     (1)\r\n" );
document.write( "(L+2*3)*(W+2*3) - L*W = 402.    (2)\r\n" );
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document.write( "Simplify (2). It is equivalent to\r\n" );
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document.write( "(L+6)*(W+6) - LW = 402,\r\n" );
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document.write( "LW + 6L + 6W + 36 - LW = 402,\r\n" );
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document.write( "6L + 6W = 402 - 36 = 366,\r\n" );
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document.write( "6*(L+W) = 366  ====>  L + W =  = 61.\r\n" );
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document.write( "So, the problem is equivalent to these two equations\r\n" );
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document.write( "L - W = 10,\r\n" );
document.write( "L + W = 61.\r\n" );
document.write( "------------------Add the equations (both sides). You will get\r\n" );
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document.write( "2L = 10 + 61 = 71  ====>  L =  = 35.5.\r\n" );
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document.write( "Answer.  The dimensions of the oil painting are 35.5 inches (length) and 25.5 inches (width).\r\n" );
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