document.write( "Question 683587: A person invested $20,000, part at an interest rate of 6% annually and the remainder at 7% annually. The total interest at the end of 1 year was equivalent to an annual 6.75% rate on the entire $20,000. How much was invested at each rate?
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Algebra.Com's Answer #423623 by mananth(16949)\"\" \"About 
You can put this solution on YOUR website!
6.75% on 20,000 = 1350 total interest earned.\r
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\n" ); document.write( "\n" ); document.write( "Part I 6.00% per annum ------------- Amount invested =x
\n" ); document.write( "Part II 7.00% per annum ------------ Amount invested = y
\n" ); document.write( " 20000
\n" ); document.write( "Interest----- 1350
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\n" ); document.write( "Part I 6.00% per annum ---x
\n" ); document.write( "Part II 7.00% per annum ---y
\n" ); document.write( "Total investment
\n" ); document.write( "x + 1 y= 20000 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "6.00% x + 7.00% y= 1350
\n" ); document.write( "Multiply by 100
\n" ); document.write( "6 x + 7 y= 135000.00 --------2
\n" ); document.write( "Multiply (1) by -6
\n" ); document.write( "we get
\n" ); document.write( "-6 x -6 y= -120000.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x 1 y= 15000
\n" ); document.write( "divide by 1
\n" ); document.write( " y = 15000
\n" ); document.write( "Part I 6.00% $ 5000
\n" ); document.write( "Part II 7.00% $ 15000
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\n" ); document.write( "CHECK
\n" ); document.write( "5000 --------- 6.00% ------- 300.00
\n" ); document.write( "15000 ------------- 7.00% ------- 1050.00
\n" ); document.write( "Total -------------------- 1350.00
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\n" ); document.write( "m.ananth@hotmail.ca
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