document.write( "Question 708692:
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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%? \n" );
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Algebra.Com's Answer #436234 by checkley79(3341)![]() ![]() ![]() You can put this solution on YOUR website! .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 \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |