document.write( "Question 1107868: Mary invested a total of ​$48,000 in three different bank accounts. One account pays an annual interest rate of 2​%, the second account pays 3​% annual​ interest, and the third account pays 4​% annual interest. In one​ year, Mary earned a total of ​$1,550 in interest from these three accounts. If Mary invested ​$8,000 more in the account that pays 3​% interest than she did in the account that pays 4​% ​interest, find the amount she invested in each account. \n" ); document.write( "
Algebra.Com's Answer #722896 by Boreal(15235)\"\" \"About 
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x is 2%
\n" ); document.write( "y=3%
\n" ); document.write( "z=4%
\n" ); document.write( "but z=y-8000
\n" ); document.write( "Therefore,
\n" ); document.write( "x+y+y-8000=48000
\n" ); document.write( "x+2y=56000
\n" ); document.write( ".02x+.03y+.04(y-8000)=1550
\n" ); document.write( ".02x+.03y+.04y-320=1550
\n" ); document.write( ".02x+.07y=1870
\n" ); document.write( "x+2y=56000
\n" ); document.write( "multiply the top by -50
\n" ); document.write( "-x-3.5y=-93500
\n" ); document.write( "add
\n" ); document.write( "-1.5y=-37500
\n" ); document.write( "y=25000@3%=$750
\n" ); document.write( "y-8000=17000@4%=$680
\n" ); document.write( "remaining is x
\n" ); document.write( "6000@0.02 for $120\r
\n" ); document.write( "\n" ); document.write( "invested
\n" ); document.write( "$6000 at 2%
\n" ); document.write( "$25000 at 3%
\n" ); document.write( "$17000 at 4%
\n" ); document.write( " \n" ); document.write( "
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