document.write( "Question 426469: Deshi invested $32,000. He invested part of it at 8% interest per year and the rest at 6% interest per year. If his total interest income for the year was $2120, ow much was invested at 6%? \n" ); document.write( "
Algebra.Com's Answer #296674 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Part I 6.00% per annum \n" ); document.write( "Part II 8.00% per annum \n" ); document.write( " \n" ); document.write( "x+y=32000 ------------------------1 \n" ); document.write( "6.00%x+8.00%y=$2,120.00 \n" ); document.write( "Multiply by 100 \n" ); document.write( "6x+8y=$212,000.00 --------2 \n" ); document.write( " \n" ); document.write( "Multiply (1) by -6 \n" ); document.write( "we get \n" ); document.write( "-6x-6y= -192000 \n" ); document.write( "Add this to (2) \n" ); document.write( "2y=$20,000.00 \n" ); document.write( " \n" ); document.write( "divide by 2 \n" ); document.write( "investment at 8.00% $10,000 \n" ); document.write( "Balance $22,000.00 investment at 6.00% \n" ); document.write( " \n" ); document.write( " |