document.write( "Question 422133: A student invested $5000 in two different savings accounts. The first account pays an annual interest rate of 3%. The second account pays an annual interest rate of 4%. At the end of the year, she had earned $185 in interest. How much money did she invest in each account? \n" ); document.write( "
Algebra.Com's Answer #294688 by mananth(16946)![]() ![]() You can put this solution on YOUR website! 3.00% per annum ------------- x \n" ); document.write( "4.00% per annum ------------ y \n" ); document.write( "total investment 5000 \n" ); document.write( "Interest----- $185.00 \n" ); document.write( "... \n" ); document.write( "x+y=5000 ------------------------1 \n" ); document.write( "3.00%x+4.00%y=$185.00 \n" ); document.write( "Multiply by 100 \n" ); document.write( "3x+4y=$18,500.00 --------2 \n" ); document.write( " \n" ); document.write( "Multiply (1) by -3 \n" ); document.write( "we get \n" ); document.write( "-3x-3y= -15000 \n" ); document.write( " \n" ); document.write( "Add this to (2) \n" ); document.write( " \n" ); document.write( "y=$3,500.00 investment at 4% \n" ); document.write( "Balance $1,500.00 investment at 3% \n" ); document.write( " \n" ); document.write( " |