document.write( "Question 872358: A man has $10,000 to invest, some in a relatively safe account
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document.write( "earning 5% interest per year, and the rest in more speculative
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document.write( "investments earning 12% per year. If the total interest earned for the
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document.write( "year was $955, how much was invested at each rate?
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Algebra.Com's Answer #526094 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Part I 5.00% per annum ------------- Amount invested =x \n" ); document.write( "Part II 12.00% per annum ------------ Amount invested = y \n" ); document.write( " 10000 \n" ); document.write( "Interest----- 955.00 \n" ); document.write( " \n" ); document.write( "Part I 5.00% per annum ---x \n" ); document.write( "Part II 12.00% per annum ---y \n" ); document.write( "Total investment \n" ); document.write( "x + 1 y= 10000 -------------1 \n" ); document.write( "Interest on both investments \n" ); document.write( "5.00% x + 12.00% y= 955 \n" ); document.write( "Multiply by 100 \n" ); document.write( "5 x + 12 y= 95500.00 --------2 \n" ); document.write( "Multiply (1) by -5 \n" ); document.write( "we get \n" ); document.write( "-5 x -5 y= -50000.00 \n" ); document.write( "Add this to (2) \n" ); document.write( "0 x 7 y= 45500 \n" ); document.write( "divide by 7 \n" ); document.write( " y = 6500 \n" ); document.write( "Part I 5.00% $ 3500 \n" ); document.write( "Part II 12.00% $ 6500 \n" ); document.write( " \n" ); document.write( "CHECK \n" ); document.write( "3500 --------- 5.00% ------- 175.00 \n" ); document.write( "6500 ------------- 12.00% ------- 780.00 \n" ); document.write( "Total -------------------- 955.00 \n" ); document.write( " \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " \n" ); document.write( " |