document.write( "Question 178291: I need some help with this problem. I have worked it several diffrent ways and can not come up with the answer.\r
\n" ); document.write( "\n" ); document.write( "If $2000 is invested at 10% simple annual interest, how much should be invested at 12% annual simple interest so that the total yearly income from both investments is $5000?\r
\n" ); document.write( "\n" ); document.write( "THis is the way I tried it
\n" ); document.write( ".12x+.1(2000-x)=5000
\n" ); document.write( ".12x+200-.1x=5000
\n" ); document.write( ".02x+200=5000
\n" ); document.write( " -200 -200
\n" ); document.write( ".02x=4800
\n" ); document.write( "x=240,000\r
\n" ); document.write( "\n" ); document.write( "I know that is not the correct answer because of the value.
\n" ); document.write( "Any help with this is greatly appreciated!! \n" ); document.write( "

Algebra.Com's Answer #133263 by stanbon(75887) $\"\"$ $\"About$

You can put this solution on YOUR website!
If $2000 is invested at 10% simple annual interest, how much should be invested at 12% annual simple interest so that the total yearly income from both investments is $5000?
\n" ); document.write( "=====================
\n" ); document.write( "Equation:
\n" ); document.write( "interest + interest = interext
\n" ); document.write( "0.10*2000 + 0.12x = 5000\r
\n" ); document.write( "\n" ); document.write( "Multiply thru by 100 to get:
\n" ); document.write( "10*2000 + 12x = 500000
\n" ); document.write( "12x = 480000
\n" ); document.write( "x - $40,000 (amt. that must be invested at 12%)
\n" ); document.write( "--------------
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "

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