document.write( "Question 113986: Bob invested $20,000, part at 14% and part at 13%. If the total interest at the end of the year is $2,720, how much did he invest at 14%? \n" ); document.write( "
Algebra.Com's Answer #82944 by checkley71(8403)![]() ![]() ![]() You can put this solution on YOUR website! .14X+.13(20,000-X)=2,720 \n" ); document.write( ".14X+2600-.13X=2,720 \n" ); document.write( ".01X=2,720-2,600 \n" ); document.write( ".01X=120 \n" ); document.write( "X=120/.01 \n" ); document.write( "X=$12,000 IS THE AMOUNT INVESTED @ 14%. \n" ); document.write( "PROOF: \n" ); document.write( ".14*12,000+.13(20,000-12,000)=2,720 \n" ); document.write( "1,680+.13*8,000=2,720 \n" ); document.write( "1,680+1,040=2,720 \n" ); document.write( "2,720=2,720 \n" ); document.write( " |