document.write( "Question 301000: the perimeter of a rectangle is 64 cm. the length is 4 less than twice the width. find the length and width \n" ); document.write( "

Algebra.Com's Answer #215860 by Deina(147) $\"\"$ $\"About$

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Variables Used

p = perimeter

l = length

w = width

\n" ); document.write( "What do we know? We know that:
\n" ); document.write( "p = 64 cm
\n" ); document.write( "l = 2w - 4
\n" ); document.write( "and finally we know that
\n" ); document.write( "p = 2(l + w) or 2l + 2w

\n" ); document.write( "so we will substitute 2w - 4 for l in the formula:
\n" ); document.write( "p = 2(2w - 4) + 2w
\n" ); document.write( "Now we will distribute:
\n" ); document.write( "p = 4w - 8 + 2w
\n" ); document.write( "combine like terms:
\n" ); document.write( "p = 6w - 8
\n" ); document.write( "I like working on the left, so we'll flip-flop (advanced mathematical term) and replace \"p\" with 64:
\n" ); document.write( "6w - 8 = 64

\n" ); document.write( "Add 8 to each side:
\n" ); document.write( "6w = 72
\n" ); document.write( "divide each side by 6:
\n" ); document.write( " $\"+cross%286%29w%2Fcross%286%29+=+72%2F6+=+12+\"$ so
\n" ); document.write( "w = 12
\n" ); document.write( "now remember that l = 2w - 4, and substitute 12 for 'w'
\n" ); document.write( " $\"+l+=+cross%282%2812%29%29+24+-+4+=+20+\"$
\n" ); document.write( "l = 20\r
\n" ); document.write( "\n" ); document.write( "Your answers are
\n" ); document.write( "Length = 20 cm
\n" ); document.write( "Width = 12 cm & bob's your uncle!\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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