document.write( "Question 351137: This is a mathematics problem which one of my cousin asked me yesterday. He is in grade 4. However, I cannot help him because I am not good in maths.\r
\n" ); document.write( "\n" ); document.write( "The width of a rectangle is 20 cm more than half the length.
\n" ); document.write( "Its area is 4800 sq.cm.
\n" ); document.write( "Find its width.
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Algebra.Com's Answer #250981 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Start with the formula for the area of a rectangle:
\n" ); document.write( "\"A+=+L%2AW\" Where L = length and W = width.
\n" ); document.write( "The area, A, is given as A = 4800 sq.cm.
\n" ); document.write( "The width, W, is given as:
\n" ); document.write( "\"W+=+%281%2F2%29L%2B20\" \"The width ... is 20cm more than half the length.\"
\n" ); document.write( "Let's express the above in terms of the length, L.
\n" ); document.write( "\"W+=+%281%2F2%29L%2B20\" Multiply through by 2 to clear the fraction.
\n" ); document.write( "\"2W+=+L%2B40\" Subtract 40 from both sides.
\n" ); document.write( "\"2W-40+=+L\" Now substitute this into the formula for L.
\n" ); document.write( "\"A+=+L%2AW\" Substitute A = 4800, L = 2W-40
\n" ); document.write( "\"4800+=+%282W-40%29%2AW\" Simplify.
\n" ); document.write( "\"4800+=+2W%5E2-40W\" Now subtract 4800 from both sides.
\n" ); document.write( "\"2W%5E2-40W-4800+=+0\" Divide by 2 to simplify a bit.
\n" ); document.write( "\"W%5E2-20W-2400+=+0\" Factor this quadratic equation.
\n" ); document.write( "\"%28W%2B40%29%28W-60%29+=+0\" Apply the zero product rule.
\n" ); document.write( "\"W%2B40+=+0\" or \"W-60+=+0\" so...
\n" ); document.write( "\"W+=+-40\" or \"W+=+60\" Discard the negative solution as width is a positive quantity.
\n" ); document.write( "The width is 60cm.
\n" ); document.write( "Note: This is certainly well beyond grade 4 level math in american schools.
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