document.write( "Question 779591: A total of $3700 was invested, part of it at 6% interest and the remainder at 11%. If the total yearly interest amounted to $330, how much was invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #475282 by mananth(16946)\"\" \"About 
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Part I 6.00% per annum ------------- Amount invested =x
\n" ); document.write( "Part II 11.00% per annum ------------ Amount invested = y
\n" ); document.write( " 3700
\n" ); document.write( "Interest----- 330.00
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\n" ); document.write( "Part I 6.00% per annum ---x
\n" ); document.write( "Part II 11.00% per annum ---y
\n" ); document.write( "Total investment
\n" ); document.write( "x + 1 y= 3700 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "6.00% x + 11.00% y= 330
\n" ); document.write( "Multiply by 100
\n" ); document.write( "6 x + 11 y= 33000.00 --------2
\n" ); document.write( "Multiply (1) by -6
\n" ); document.write( "we get
\n" ); document.write( "-6 x -6 y= -22200.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x 5 y= 10800
\n" ); document.write( "divide by 5
\n" ); document.write( " y = 2160 - x)
\n" ); document.write( "Part I 6.00% $ 1540
\n" ); document.write( "Part II 11.00% $ 2160
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "1540 --------- 6.00% ------- 92.40
\n" ); document.write( "2160 ------------- 11.00% ------- 237.60
\n" ); document.write( "Total -------------------- 330.00
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\n" ); document.write( "m.ananth@hotmail.ca
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