document.write( "Question 222411: Two investments were made totaling $3,000. For a certain year, these investments yielded $216 in simple interest. Part of the $3,000 was invested at 6% and part at 8%. Find the amount invested at each rate. \n" ); document.write( "
Algebra.Com's Answer #166517 by rfer(16322)![]() ![]() ![]() You can put this solution on YOUR website! .06x+.08(3000-x)=216 \n" ); document.write( ".06x+240-.08x=216 \n" ); document.write( "-.02x=-24 \n" ); document.write( "x=$1200 at 6% \n" ); document.write( "3000-x=$1800 at 8% \n" ); document.write( " |