document.write( "Question 351052: 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 #250916 by mananth(16946)\"\" \"About 
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$26,000 total investment
\n" ); document.write( "6% ------------$x
\n" ); document.write( "3%, ------------26000-x
\n" ); document.write( "interest =$1,080,
\n" ); document.write( "...
\n" ); document.write( "0.06x+0.03(26000-x)=1080
\n" ); document.write( "multiply by 100
\n" ); document.write( "6x +3(26000-x)=108000
\n" ); document.write( "6x+78000-3x=108000
\n" ); document.write( "3x=108000-78000
\n" ); document.write( "3x=30000
\n" ); document.write( "x= $10000 @ 6%
\n" ); document.write( "$16000 @ 3%
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