document.write( "Question 731962: \r
\n" ); document.write( "\n" ); document.write( "Marissa invested $7000 for 1 year. Part was invested at 4% simple interest and part at 7% simple interest. At the end of 1 year, she had earned $391 in interest. How much had she invested at each rate?
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Algebra.Com's Answer #447507 by checkley79(3341)\"\" \"About 
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.07X+.04(7000-X)=391
\n" ); document.write( ".07X+280-.04X=391
\n" ); document.write( ".03X=391-280
\n" ); document.write( ".03X=111
\n" ); document.write( "X=111/.03
\n" ); document.write( "X=$3,700 WAS INVESTED @ 7%
\n" ); document.write( "7000-3700=$3300 WAS INVESTED @ 4%
\n" ); document.write( "PROOF:
\n" ); document.write( ".07*3700+.04*3300=391
\n" ); document.write( "259+132=391
\n" ); document.write( "391=391
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