document.write( "Question 923226: One of the accounts pay 9% annual interest, whereas the other pays 11% annual interest. If you have $700 more invested at 11% than you invested at 9%, how much do you have invested in each account if the total amount of interest you earn in a year is $137?
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Algebra.Com's Answer #559912 by richwmiller(17219)\"\" \"About 
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We don't know the total invested.
\n" ); document.write( "We know the difference in the two accounts.
\n" ); document.write( "We know that the interest for the two accounts is $137
\n" ); document.write( "0.11*x+0.09*y=137
\n" ); document.write( "We know that the account at 11% has $700 more.
\n" ); document.write( "x=700+y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.11*(700+y)+0.09*y=137
\n" ); document.write( "We multiply out
\n" ); document.write( "77+0.11y+0.09*y=137
\n" ); document.write( "We combine like terms.
\n" ); document.write( "0.2*y=60
\n" ); document.write( "Isolate y
\n" ); document.write( "y=$300 at 9%
\n" ); document.write( "x=700+y
\n" ); document.write( "Calculate x
\n" ); document.write( "x=$1000 at 11%
\n" ); document.write( "Now,we know the total invested is: 1300
\n" ); document.write( "Total invested $1000+$300=$1300
\n" ); document.write( "We check
\n" ); document.write( "0.11*1000+0.09*300=137
\n" ); document.write( "110+27=137
\n" ); document.write( "137=137
\n" ); document.write( "Since this statement is TRUE and neither x nor y is negative all is well.
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