document.write( "Question 1126297: Can you solve for L Please\r
\n" ); document.write( "\n" ); document.write( "P=2L+2W+2H
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Algebra.Com's Answer #742614 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Answer: \"L+=+%28P-2W-2H%29%2F2\"\r
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\n" ); document.write( "Work Shown:\r
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\n" ); document.write( "\n" ); document.write( "We want to solve for L, so we have to get it all by itself on one side. We want to isolate this term. \r
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\n" ); document.write( "\n" ); document.write( "The way we do that is to undo PEMDAS. We follow the order of operations in reverse. So we'll undo subtraction first (if there is any subtraction going on), then undo addition (if there is any addition going on) etc.\r
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\n" ); document.write( "\n" ); document.write( "\"P=2L%2B2W%2B2H\" Here is the starting equation we are given\r
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\n" ); document.write( "\n" ); document.write( "\"P-2W=2L%2B2W%2B2H-2W\" Undoing addition; Subtract 2W from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"P-2W=2L%2B2H\" The 2W term goes away on the right side\r
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\n" ); document.write( "\n" ); document.write( "\"P-2W-2H=2L%2B2H-2H\" Undoing addition; Subtract 2H from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"P-2W-2H=2L\" The 2H term goes away on the right side\r
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\n" ); document.write( "\n" ); document.write( "\"2L+=+P-2W-2H\" Flip the equation\r
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\n" ); document.write( "\n" ); document.write( "\"%282L%29%2F2+=+%28P-2W-2H%29%2F2\" Undoing multiplication; divide both sides by 2\r
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\n" ); document.write( "\n" ); document.write( "\"L+=+%28P-2W-2H%29%2F2\" Simplify the left side\r
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\n" ); document.write( "\n" ); document.write( "Optionally we could break up the fraction on the right side to get this
\n" ); document.write( "\"L+=+%28P-2W-2H%29%2F2\"
\n" ); document.write( "\"L+=+P%2F2-%282W%29%2F2-%282H%29%2F2\"
\n" ); document.write( "\"L+=+P%2F2-W-H\"
\n" ); document.write( "but again it's optional
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