document.write( "Question 7850: I've been working on this for about 15 minutes now and I can not seem to find an answer. Please help.\r
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\n" ); document.write( "\n" ); document.write( " The Murrays have invested money in two accounts. One pays 7% interest and the other pays 8.5% interest. They invested $1000 more at 8.5% than they invested at 7%. If the total interest earned in one year is $395, how much money have they invested in each account.
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Algebra.Com's Answer #4306 by Earlsdon(6294)\"\" \"About 
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Let x = amount invested at 7%, then x+$1,000 = the amount invested at 8.5%.
\n" ); document.write( "Total interest earned = $395.00. Change the percents to decimals.\r
\n" ); document.write( "\n" ); document.write( "So, \"0.07x+%2B+0.085%28x%2B1000%29+=+395\"\r
\n" ); document.write( "\n" ); document.write( "\"0.07x+%2B+0.085x+%2B+85+=+395\" Simplify and solve for x.\r
\n" ); document.write( "\n" ); document.write( "\"0.155x+=+310\"\r
\n" ); document.write( "\n" ); document.write( "\"x+=+2000\"\r
\n" ); document.write( "\n" ); document.write( "So the Murrays invested $2,000.00 at 7% interest and $3,000.00 at 8.5% interest.\r
\n" ); document.write( "\n" ); document.write( "Check: \r
\n" ); document.write( "\n" ); document.write( "0.07($2,000.00) + 0.085($3,000.00) = $140.00 + $255.00 = $395.00\r
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