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) You can put this solution on YOUR website! $26,000 total investment \n" ); document.write( "6% ------------$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% \n" ); document.write( " |