document.write( "Question 667741: A woman has $450 invested, part at 2% and the remainder at 3% simple interest. How much is invested at each rate if the total annual income from these investment is $11?\r
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Algebra.Com's Answer #415150 by mananth(16946)![]() ![]() You can put this solution on YOUR website! Part I 2.00% per annum ------------- Amount invested =x \n" ); document.write( "Part II 3.00% per annum ------------ Amount invested = y \n" ); document.write( " 450 \n" ); document.write( "Interest----- 11 \n" ); document.write( " \n" ); document.write( "Part I 2.00% per annum ---x \n" ); document.write( "Part II 3.00% per annum ---y \n" ); document.write( "Total investment \n" ); document.write( "x + y= 450 -------------1 \n" ); document.write( "Interest on both inveents \n" ); document.write( "2.00% x + 3.00% y= 11 \n" ); document.write( "Multiply by 2 \n" ); document.write( "2 x + 3 y= 1100.00 --------2 \n" ); document.write( "Multiply (1) by -2 \n" ); document.write( "we get \n" ); document.write( "-2 x -2 y= -900.00 \n" ); document.write( "Add this to (2) \n" ); document.write( "0 x 1 y= 200 \n" ); document.write( "divide by 1 \n" ); document.write( " y = 200 \n" ); document.write( "Part I 2.00% $ 250 \n" ); document.write( "Part II 3.00% $ 200 \n" ); document.write( " \n" ); document.write( "CHECK \n" ); document.write( "250 --------- 2.00% ------- 5.00 \n" ); document.write( "200 ------------- 3.00% ------- 6.00 \n" ); document.write( "Total -------------------- 11.00 \n" ); document.write( " \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " \n" ); document.write( " |