document.write( "Question 708692:
\n" ); document.write( "An investment club invested part of $10,000 at 4% annual interest and the rest at 8%. If the annual income from these investments was $675, how much was invested at 8%?
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Algebra.Com's Answer #436234 by checkley79(3341)\"\" \"About 
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.08X+.04(10,000-X)=675
\n" ); document.write( ".08X+400-.04X=675
\n" ); document.write( ".04X=675-400
\n" ); document.write( ".04X=275
\n" ); document.write( "X=275/.04
\n" ); document.write( "X=$6,875 AMOUNT INVESTED @ 8%.
\n" ); document.write( "10,000-6875=$3,125 AMOUNT INVESTED @ 4%.
\n" ); document.write( "PROOF:
\n" ); document.write( ".08*6875+.04*3125=675
\n" ); document.write( "550+125=675
\n" ); document.write( "675=675\r
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