document.write( "Question 1102815: Susan invests
\n" ); document.write( "3 times as much money at \r
\n" ); document.write( "\n" ); document.write( "11% as she does at \r
\n" ); document.write( "\n" ); document.write( "7%. If her total interest after
\n" ); document.write( "1 year is \r
\n" ); document.write( "\n" ); document.write( "$2400, how much does she have invested at each rate?
\n" ); document.write( "

Algebra.Com's Answer #717553 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "Let x be the amount invested at 7%; then the amount invested at 11% is 3x.

\n" ); document.write( "The interest from the first investment is then .07 times x; the interest from the second is .11 times 3x.

\n" ); document.write( "Since the total interest is $2400,
\n" ); document.write( "\".07%28x%29%2B.11%283x%29+=+2400\"
\n" ); document.write( "\".07x%2B.33x+=+2400\"
\n" ); document.write( "\".40x+=+2400\"
\n" ); document.write( "\"x+=+2400%2F.40+=+6000\"

\n" ); document.write( "She invested x = $6000 at 7% and 3x = $18000 at 11%.

\n" ); document.write( "CHECK:
\n" ); document.write( "\".07%286000%29%2B.11%2818000%29+=+420%2B1980+=+2400\"
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