document.write( "Question 269869: Pat invested a total of $3,000. Part of the money yields 10 percent interest per year and the rest yields 8 percent interest per year. If the total yearly interest from this investment is $256, how much did Pat invest at 10 percent and how much at 8 percent? \n" ); document.write( "
Algebra.Com's Answer #197705 by checkley77(12844) ![]() You can put this solution on YOUR website! .10x+.08(3,000-x)=256 \n" ); document.write( ".10x+240-.08x=256 \n" ); document.write( ".02x=256-240 \n" ); document.write( ".02x=16 \n" ); document.write( "x=16/.02 \n" ); document.write( "x=$800 amount invested @ 10%. \n" ); document.write( "3,000-800=$2,200 amount invested @ 8%. \n" ); document.write( "Proof: \n" ); document.write( ".10*800+.08*2,200=256 \n" ); document.write( "80+176=256 \n" ); document.write( "256=256\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |