document.write( "Question 203687: Janet invested $26,000, part at 6% and part at 3%. If the total interest at the end of the year is 1,080, how much did she invest at 6%? \n" ); document.write( "
Algebra.Com's Answer #153682 by rfer(16322)\"\" \"About 
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let x = 6% part= $10000
\n" ); document.write( "let 26000-x= 3% part= $16000
\n" ); document.write( ".06x+.03(26000-x)=1080
\n" ); document.write( ".06x+780-.03x=1080
\n" ); document.write( ".03x=300
\n" ); document.write( "x=10000
\n" ); document.write( "
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