document.write( "Question 899358: Juanita has 22000 to invest in two accounts that pay simple interest on an annual basis. one account pays 3% in simple interest, and other pays 6% simple interest how much would she have to invest in each account to earn a total interest of $810 after one year? \n" ); document.write( "
Algebra.Com's Answer #545329 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Part I 3.00% per annum ------------- Amount invested =x \n" ); document.write( "Part II 6.00% per annum ------------ Amount invested = y \n" ); document.write( " 22000 \n" ); document.write( "Interest----- 810.00 \n" ); document.write( " \n" ); document.write( "Part I 3.00% per annum ---x \n" ); document.write( "Part II 6.00% per annum ---y \n" ); document.write( "Total investment \n" ); document.write( "x + 1 y= 22000 -------------1 \n" ); document.write( "Interest on both investments \n" ); document.write( "3.00% x + 6.00% y= 810 \n" ); document.write( "Multiply by 100 \n" ); document.write( "3 x + 6 y= 81000.00 --------2 \n" ); document.write( "Multiply (1) by -3 \n" ); document.write( "we get \n" ); document.write( "-3 x -3 y= -66000.00 \n" ); document.write( "Add this to (2) \n" ); document.write( "0 x 3 y= 15000 \n" ); document.write( "divide by 3 \n" ); document.write( " y = 5000 \n" ); document.write( "Part I 3.00% $ 17000 \n" ); document.write( "Part II 6.00% $ 5000 \n" ); document.write( " \n" ); document.write( "CHECK \n" ); document.write( "17000 --------- 3.00% ------- 510.00 \n" ); document.write( "5000 ------------- 6.00% ------- 300.00 \n" ); document.write( "Total -------------------- 810.00 \n" ); document.write( " \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " \n" ); document.write( " |