document.write( "Question 5892: Suppose jodi invested $10,000 , part at 6% simple annual interest, and the rest at 9% simple annual interest. If you received $684 in simple interest after one year, how much did you invest at each rate?\r
\n" ); document.write( "\n" ); document.write( "My answer was:
\n" ); document.write( "$3999.60 at 6%
\n" ); document.write( "$5999.40 at 9%
\n" ); document.write( "Is this the right answer? please help me! \n" ); document.write( "
Algebra.Com's Answer #3073 by Abbey(339)\"\" \"About 
You can put this solution on YOUR website!
You can check your answer by multiplication:
\n" ); document.write( ".06*3999.60 = 239.976
\n" ); document.write( ".09*5999.40 = 539.946
\n" ); document.write( "239.976+539.946 = 779.922 - not a correct answer because it should only total $684:\r
\n" ); document.write( "\n" ); document.write( "Let x = the amount jodi invested at 6%
\n" ); document.write( ".06x = the interest earned at 6%
\n" ); document.write( "Let 10000-x = the amount invested at 9% ($10000-the amount invested at 6%=the amount invested at 9%)
\n" ); document.write( ".08(10000-x)= the interest earned at 9%
\n" ); document.write( "6% interest + 9% interest = $684
\n" ); document.write( ".06x+.09(10000-x)=684
\n" ); document.write( ".06x+900-.08x=684
\n" ); document.write( "-.03x=684-900
\n" ); document.write( "-.03x=-216
\n" ); document.write( "x=7200
\n" ); document.write( "amount invested at 6%=7200
\n" ); document.write( "amount invested at 9% = 2800 (10000-7200)\r
\n" ); document.write( "\n" ); document.write( "check:
\n" ); document.write( ".06*7200=$432
\n" ); document.write( ".09*2800 = $252
\n" ); document.write( "432+252=$684\r
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