document.write( "Question 824314: I need more detail solution:\r
\n" ); document.write( "\n" ); document.write( "A girl has orange, yellow and red sweets. She has exactly twice as many red than yellow sweets.
\n" ); document.write( "After eating fourteen orange sweets, she has two less orange than yellow sweets. Also, the
\n" ); document.write( "number of orange sweets now are 20% of the total number of sweets she had at the beginning.
\n" ); document.write( "How many sweets did she have at the beginning?
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Algebra.Com's Answer #496307 by stanbon(75887)\"\" \"About 
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I need more detail solution:
\n" ); document.write( "A girl has orange, yellow and red sweets.
\n" ); document.write( "number = o + y + r
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\n" ); document.write( "She has exactly twice as many red as yellow.
\n" ); document.write( "number = o + y + 2y = o+3y
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\n" ); document.write( "After eating fourteen orange sweets, she has two less orange than yellow sweets.
\n" ); document.write( "o-14 = y-2
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\n" ); document.write( "Also, the number of orange sweets now are 20% of the total number of sweets she had at the beginning.
\n" ); document.write( "o-14 = (1/5)(o+3y)
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\n" ); document.write( "How many sweets did she have at the beginning?
\n" ); document.write( "Using o-14 = (1/5)(o+3y) and o-14 = y-2 you get::
\n" ); document.write( "5o-70 = o+3y
\n" ); document.write( "4o - 3y = 70
\n" ); document.write( "o = y+12
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\n" ); document.write( "Solve for \"y\":
\n" ); document.write( "4y+48-3y = 70
\n" ); document.write( "y = 22 (# of yellow sweets)
\n" ); document.write( "o = y + 12 = 34 (# of orange sweets)
\n" ); document.write( "r = 2y = 44 (# of red sweets)
\n" ); document.write( "----
\n" ); document.write( "total number:: 22 + 34 + 44 = 100
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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