document.write( "Question 102343: distibute 1000 so that one part has 3 times as large as the second one and the third has the sum of the other two \n" ); document.write( "
Algebra.Com's Answer #74448 by Fombitz(32388)\"\" \"About 
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First Part + Second Part + Third Part = 1000
\n" ); document.write( "First Part = 3 x Second Part
\n" ); document.write( "Third Part = First Part + Second Part\r
\n" ); document.write( "\n" ); document.write( "Let's assign a variable to each part.
\n" ); document.write( "X = First Part
\n" ); document.write( "Y = Second Part
\n" ); document.write( "Z = Third Part\r
\n" ); document.write( "\n" ); document.write( "\"X+%2B+Y+%2B+Z+=+1000\" Your original equation.
\n" ); document.write( "\"X+=+3%2AY\" First Part = 3 x Second Part
\n" ); document.write( "\"Z+=+X+%2B+Y+=+%283%2AY%29+%2B+Y+=+4Y\"Third Part = First Part + Second Part
\n" ); document.write( "\"X+%2B+Y+%2B+Z+=+1000\"
\n" ); document.write( "\"3%2AY+%2B+Y+%2B+4%2AY+=+1000\" Substitute for what you know, in terms of Y. \
\n" ); document.write( "\"8%2AY+=+1000\" Simplify.
\n" ); document.write( "\"cross%288%29Y%2Fcross%288%29+=+1000%2F8\" Use multiplicative inverse of 8.
\n" ); document.write( "\"Y+=+125\"
\n" ); document.write( "\"X+=+3%2AY+=+3%2A%28125%29+=+375\"
\n" ); document.write( "\"Z+=+X+%2B+Y+=+125+%2B+375+=+500\"
\n" ); document.write( "So then,
\n" ); document.write( "First Part = 125
\n" ); document.write( "Second Part = 375
\n" ); document.write( "Third Part = 500
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