document.write( "Question 840868: An executive invests $24,000, some at 6% and some at 5% annual interest. If he receives an annual return of $1,380, how much is invested at each rate? \n" ); document.write( "
Algebra.Com's Answer #506444 by hamsanash1981@gmail.com(151)\"\" \"About 
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Let the amount invested at 6% and 5% annual interest be x and y. the number of years be 1.\r
\n" ); document.write( "\n" ); document.write( "the x + Y = 24000 ---- 1
\n" ); document.write( "and simple interest for x is (x*1*6)/100 => 6X/100
\n" ); document.write( "simple interest for y is (y*1*5)/100 => 5y/100
\n" ); document.write( "Given that he recieves an annual returns of $1,380, then
\n" ); document.write( "6x/100 + 5y/100 = 1380 --- 2\r
\n" ); document.write( "\n" ); document.write( "multipling equation 2 by 100\r
\n" ); document.write( "\n" ); document.write( "6x+ 5y = 138000 --- 3\r
\n" ); document.write( "\n" ); document.write( "Solving equ 1 and 3\r
\n" ); document.write( "\n" ); document.write( "1*6 6x + 6y = 144000
\n" ); document.write( " 6x + 5y = 138000
\n" ); document.write( " ----------------
\n" ); document.write( " y = 6000
\n" ); document.write( "then x = 24000 - 6000 => x = 18000\r
\n" ); document.write( "\n" ); document.write( "therefore 18000 is invested at 6% and 6000 is invested at 5%.
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