document.write( "Question 1109146: a person invests $20000, partly at 5%, partly at 6% and the remainder at 6.5%. the total annual interest is $1170. three times the amount invested at 6% is equals the amount invested at 5% and 6.5% combined. how much money is invested at each rate \n" ); document.write( "
Algebra.Com's Answer #724186 by TeachMath(96)\"\" \"About 
You can put this solution on YOUR website!
You can use just 2 variables. Using more can make the problem a lot more complex, in my opinion.\r
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\n" ); document.write( "\n" ); document.write( "Let amount invested at 5% and 6.5% be F and S, respectively
\n" ); document.write( "Then amount invested at 6% = 20,000 – F – S
\n" ); document.write( "We then get: .05F + .065S + .06(20,000 – F – S) = 1,170
\n" ); document.write( ".05F + .065S + 1,200 - .06F - .06S = 1,170
\n" ); document.write( "- .01F + .005S = - 30 ------- eq 1\r
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\n" ); document.write( "\n" ); document.write( "Also, 3(20,000 – F – S) = F + S
\n" ); document.write( "60,000 – 3F - 3S = F + S
\n" ); document.write( "F + 3F + S + 3S = 60,000
\n" ); document.write( "4F + 4S = 60,000___4(F + S) = 4(15,000)____F + S = 15,000___S = 15,000 - F ---- eq 2\r
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\n" ); document.write( "\n" ); document.write( "- .01F + .005(15,000 - F) = - 30 ------- Substituting 15,000 – F for S in eq 1
\n" ); document.write( "- .01F + 75 – .005F = - 30
\n" ); document.write( "- .01F - .005F = - 30 - 75
\n" ); document.write( "- .015F = - 105
\n" ); document.write( "F, or amount invested at 5% = - 105/- .015 = $7,000 \r
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\n" ); document.write( "\n" ); document.write( "S = 15,000 – 7,000 ------- Substituting 7,000 for F in eq 2
\n" ); document.write( "S, or amount invested at 6.5% = $8,000\r
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\n" ); document.write( "\n" ); document.write( "Amount invested at 6%: 20,000 – 7,000 – 8,000 = $5,000 \n" ); document.write( "
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