document.write( "Question 900957: A part of $6000 was invested at 6% annual interst and remaining at 7% annual interest. At the end of the year the total amount recieved was $6395. How much money was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #546346 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! x+y=6000, \n" ); document.write( "0.06*x+0.07*y=395 \n" ); document.write( "x=6000-y \n" ); document.write( "0.06*(6000-y)+0.07*y=395 \n" ); document.write( "360-0.06y+0.07*y=395 \n" ); document.write( "0.01*y=35 \n" ); document.write( "y=3500 at 7% \n" ); document.write( "x=6000-y \n" ); document.write( "x=2500 at 6% \n" ); document.write( "check \n" ); document.write( "0.06*2500+0.07*3500=395 \n" ); document.write( "150+245=395 \n" ); document.write( "395=395 \n" ); document.write( "ok \n" ); document.write( " \n" ); document.write( " |