document.write( "Question 910171: Susan invests 3 times as much money at 11% as she does at 9%. If her total interest after 1 year is $1680, how much does she have invested at each rate?
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Algebra.Com's Answer #552289 by richwmiller(17219) $\"\"$ $\"About$

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We know that the interest for the two accounts is $1680
\n" ); document.write( "0.11*x+0.09*y=1680
\n" ); document.write( "We know that the account at 11% has 3 times the amount at 9%
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.11*(3*y)+0.09*y=1680
\n" ); document.write( "We multiply out
\n" ); document.write( "(0.33+0.09)*y=1680
\n" ); document.write( "We combine like terms.
\n" ); document.write( "0.42*y=1680
\n" ); document.write( "Isolate y
\n" ); document.write( "y= 4000.00
\n" ); document.write( "y=$ 4000.00 at 9%
\n" ); document.write( "Calculate x
\n" ); document.write( "x=3*y
\n" ); document.write( "x=$12000.00 at 11%
\n" ); document.write( "Your total for x is $12000.00
\n" ); document.write( "Your total for y is $ 4000.00
\n" ); document.write( "Your total for x and y is $16000.00
\n" ); document.write( "Total invested $12000.00 + $ 4000.00=$16000.00
\n" ); document.write( "We check
\n" ); document.write( "0.11*12000.00+0.09* 4000.00=1680
\n" ); document.write( " 1320.00+ 360.00=1680
\n" ); document.write( "1680=1680
\n" ); document.write( "Since this statement is TRUE and neither x nor y is negative then all is well.
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