document.write( "Question 54711: 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 #36988 by tutorcecilia(2152)![]() ![]() You can put this solution on YOUR website! Ax+By=C [Use the standard equation of a line] \n" ); document.write( "(14%)(x)+(12%)($20,000-x)=$2,720 \n" ); document.write( ".14x+.12(20,000-x)=2720 \n" ); document.write( ".14x+2400-.12x=2720 [Solve for x] \n" ); document.write( ".14x-.12x=2720-2400 \n" ); document.write( ".02x=320 \n" ); document.write( "x=$16,000 \n" ); document.write( ". \n" ); document.write( "14%of $16,000 = (.14)(16,000)=$2,240 \n" ); document.write( "12% of ($20,000-$16,000)=.12(20,000-16,000)=$480 \n" ); document.write( "So, $2,240+$480=$2,720 \n" ); document.write( " |