document.write( "Question 1102815: Susan invests
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document.write( "3 times as much money at \r
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document.write( "11% as she does at \r
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document.write( "7%. If her total interest after
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document.write( "1 year is \r
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document.write( "$2400, how much does she have invested at each rate? \n" );
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Algebra.Com's Answer #717553 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \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( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "She invested x = $6000 at 7% and 3x = $18000 at 11%. \n" ); document.write( "CHECK: \n" ); document.write( " |