document.write( "Question 124525: The area of a rectangle is 40.if the length is 6 more than the width, find the dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #91287 by iluvbuilding429(41)\"\" \"About 
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The area of a rectangle is 40.if the length is 6 more than the width, find the dimensions of the rectangle.

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\n" ); document.write( "\n" ); document.write( "Ok, so you know that \"L=w%2B6\".

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\n" ); document.write( "\n" ); document.write( "Plug this and your known area into the formula for area of a rectangle which is:
\n" ); document.write( "\"A=L%2Aw\"

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\n" ); document.write( "\n" ); document.write( "Here it is with the values plugged in:
\n" ); document.write( "\"40=w%2B6%2Aw\"

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\n" ); document.write( "\n" ); document.write( "Multiply:
\n" ); document.write( "\"40=w%2B6w\"

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\n" ); document.write( "\n" ); document.write( "Add:
\n" ); document.write( "\"40=7w\"

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\n" ); document.write( "\n" ); document.write( "Divide both sides by 7:
\n" ); document.write( "5 5/7=w

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\n" ); document.write( "\n" ); document.write( "Just add 6 to that to find your length, which is 11 5/7.\r
\n" ); document.write( "\n" ); document.write( "And there's your answer.

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\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "40=5 5/7+6*5 5/7

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\n" ); document.write( "\n" ); document.write( "Multiply:
\n" ); document.write( "40=5 5/7+34 2/7

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\n" ); document.write( "\n" ); document.write( "Add:
\n" ); document.write( "\"40=40\"
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