document.write( "Question 867314: Bob invested $12,000, part at 2% and part at 11%. If the total interest at the end is 870. How much did he invest at 2%? \n" ); document.write( "
Algebra.Com's Answer #522927 by edjones(8007) You can put this solution on YOUR website! x=amount invested at 11% \n" ); document.write( ".11x+.02(12000-x)=870 \n" ); document.write( ".11x+240-.02x=870 \n" ); document.write( ".09x=630 \n" ); document.write( "9x=63000 \n" ); document.write( "x=7000 \n" ); document.write( "12000-7000 \n" ); document.write( "=$5000. invested at 2% \n" ); document.write( ". \n" ); document.write( "Ed \n" ); document.write( " |