document.write( "Question 785784: Jen invested $7,500 in two back accounts, one paying 4% simple interest, and the other paying 6%. If her annual interest from the two accounts is $420, how much did she invest at 6%? \n" ); document.write( "
Algebra.Com's Answer #477718 by mananth(16946)\"\" \"About 
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Part I 6.00% per annum ------------- Amount invested =x
\n" ); document.write( "Part II 4.00% per annum ------------ Amount invested = y
\n" ); document.write( " 7500
\n" ); document.write( "Interest----- 420.00
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\n" ); document.write( "Part I 6.00% per annum ---x
\n" ); document.write( "Part II 4.00% per annum ---y
\n" ); document.write( "Total investment
\n" ); document.write( "x + 1 y= 7500 -------------1
\n" ); document.write( "Interest on both investments
\n" ); document.write( "6.00% x + 4.00% y= 420
\n" ); document.write( "Multiply by 100
\n" ); document.write( "6 x + 4 y= 42000.00 --------2
\n" ); document.write( "Multiply (1) by -6
\n" ); document.write( "we get
\n" ); document.write( "-6 x -6 y= -45000.00
\n" ); document.write( "Add this to (2)
\n" ); document.write( "0 x -2 y= -3000
\n" ); document.write( "divide by -2
\n" ); document.write( " y = 1500 - x)
\n" ); document.write( "Part I 6.00% $ 6000
\n" ); document.write( "Part II 4.00% $ 1500
\n" ); document.write( "
\n" ); document.write( "CHECK
\n" ); document.write( "6000 --------- 6.00% ------- 360.00
\n" ); document.write( "1500 ------------- 4.00% ------- 60.00
\n" ); document.write( "Total -------------------- 420.00
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\n" ); document.write( "m.ananth@hotmail.ca
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