document.write( "Question 470602: a total of $12000 is invested in two stocks of, one of which returns 8% per year and the other 12% per year. How much has been invested in each stock if the income from the 8% stock is 3$ more than the income from the 12% stock?\r
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Algebra.Com's Answer #322783 by mananth(16946)\"\" \"About 
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Part I 8.00% per annum ------------- x
\n" ); document.write( "Part II 12.00% per annum ------------ y
\n" ); document.write( " total investment 12000
\n" ); document.write( " Interest-----= 8.00% x
\n" ); document.write( " Interest-----= 12.00% y
\n" ); document.write( " greater by 5
\n" ); document.write( " 8.00% x = 12.00% y + 5
\n" ); document.write( " multiply by 100
\n" ); document.write( " 8 x - -12 y = 500 --------1
\n" ); document.write( "x + y = 12000 -------- 2
\n" ); document.write( "eliminate y
\n" ); document.write( "Multiply 1 by 1
\n" ); document.write( "Multiply 2 by -12
\n" ); document.write( "we get
\n" ); document.write( "8 x +12 y = 500
\n" ); document.write( "-12 x-12 y = -144000
\n" ); document.write( "-4 x= -143500
\n" ); document.write( "/ -4
\n" ); document.write( " x= 35875
\n" ); document.write( "Plug the value of x in in 1
\n" ); document.write( "287000 -12 y = 500
\n" ); document.write( " -12 y = 500 -287000
\n" ); document.write( " -12 y = -286500
\n" ); document.write( " / -12
\n" ); document.write( " y= 23875
\n" ); document.write( "$ 35875 at 8.00%
\n" ); document.write( "$ 23875 at 12.00%
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