document.write( "Question 66157: A sum of money of $10000 is invested, partly at 4% and partly at 6%. If the yearly income from the two investments was $530. Find the amount invested at each rate. \n" ); document.write( "
Algebra.Com's Answer #46882 by ptaylor(2198)\"\" \"About 
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A sum of money of $10000 is invested, partly at 4% and partly at 6%. If the yearly income from the two investments was $530. Find the amount invested at each rate.\r
\n" ); document.write( "\n" ); document.write( "Let x= amount invested at 4%
\n" ); document.write( "Then 10,000-x= amount invested at 6%\r
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\n" ); document.write( "\n" ); document.write( "Now we are told that the interest accrued on the amount invested at 4% (.04)(x) plus the interest accrued on the amount invested at 6% (.06(10,000-x)) equals $530. So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( ".04x+.06(10000-x)=530\r
\n" ); document.write( "\n" ); document.write( ".04x+600-.06x=530 collect the x's together and subtract 600 from both sides:
\n" ); document.write( "-.02x=-600+530
\n" ); document.write( "-.02x=-70
\n" ); document.write( "x=$3500 amount invested at 4%\r
\n" ); document.write( "\n" ); document.write( "10,000-x=10,000-3500=$6500 amount invested at 6%\r
\n" ); document.write( "\n" ); document.write( "ck\r
\n" ); document.write( "\n" ); document.write( ".04(3500)+.06(6500)=530
\n" ); document.write( "140+390=530
\n" ); document.write( "$530=$530\r
\n" ); document.write( "\n" ); document.write( "Hope this helps-----ptaylor
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