document.write( "Question 474902: Dianne invests P3500, part at 3%, part at 4%. How much does she invest at each rate if she receive in one year interest of P115? \n" ); document.write( "
Algebra.Com's Answer #325674 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Investment I 3.00% per annum \n" ); document.write( "Investment II 4.00% per annum \n" ); document.write( " \n" ); document.write( "x+y=3500 ------------------------1 \n" ); document.write( "3.00%x+4.00%y=$115.00 \n" ); document.write( "Multiply by 100 \n" ); document.write( "3x+4y=$11,500.00--------2 \n" ); document.write( " \n" ); document.write( "Multiply (1) by -3 \n" ); document.write( "we get \n" ); document.write( " \n" ); document.write( "-3x-3y =-10500.00 \n" ); document.write( " \n" ); document.write( "Add this to (2) \n" ); document.write( " \n" ); document.write( "0 x +1 y = $1,000.00 \n" ); document.write( "y= P1,000.00 investment at 4.00% \n" ); document.write( "Balance P2,500.00 investment at 3.00% \n" ); document.write( " \n" ); document.write( " |