document.write( "Question 923674: The length of a rectangle is 5 ft longer than twice the width. if the perimeter is 100ft, find the length and width of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #562198 by Roseghanezadeh(16) $\"\"$ $\"About$

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Let's give the width the variable \"w\" and the length the variable \"l\". So it says that the length of a rectangle is 5 ft longer than twice the width which we can write as \"5 + 2w\". And that its perimeter = 100ft.
\n" ); document.write( "We know the formula for the perimeter of a rectangle is:
\n" ); document.write( "2l + 2w = p
\n" ); document.write( "Now we substitute:
\n" ); document.write( "2(5 + 2w) + 2w = 100
\n" ); document.write( "10 + 4w + 2w = 100
\n" ); document.write( "Now we take like terms to one side:
\n" ); document.write( "6w = 90
\n" ); document.write( "w = 15ft
\n" ); document.write( "We also knew that \"l = 5 + 2w\", so:
\n" ); document.write( "l = 5 + 2(15)
\n" ); document.write( "l = 35ft\r
\n" ); document.write( "\n" ); document.write( "Check with the formula \"2l + 2w = p\"
\n" ); document.write( "2(35) + 2(15) = 100
\n" ); document.write( "70 + 30 = 100
\n" ); document.write( "100 = 100 \n" ); document.write( "

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