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)\"\" \"About 
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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
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